Question1
The bisector of the exterior angle âˆ A of Î” ABC intersects side BC produced at D. Prove that .
Solution:
Given: ABC is a triangle; AD is the exterior bisector of âˆ A and meets BC produced at D; BA is produced to F.
To prove:
Construction: Draw CEDA to meet AB at E.
Proof: In Î” ABC, CE  AD cut by AC.
âˆ CAD = âˆ ACE (Alternate angles)
Similarly CE  AD cut by AB
âˆ FAD = âˆ AEC (corresponding angles)
Since âˆ FAD = âˆ CAD (given)
âˆ´ âˆ ACE = âˆ AEC
\ AC = AE ( by isosceles Î” theorem)
Now in Î” BAD, CE  DA
AE = DC (BPT)
AB BD
But AC = AE (proved above)
\ AC = DC
AB BD or
Given: ABC is a triangle; AD is the exterior bisector of âˆ A and meets BC produced at D; BA is produced to F.
To prove:
Construction: Draw CEDA to meet AB at E.
Proof: In Î” ABC, CE  AD cut by AC.
âˆ CAD = âˆ ACE (Alternate angles)
Similarly CE  AD cut by AB
âˆ FAD = âˆ AEC (corresponding angles)
Since âˆ FAD = âˆ CAD (given)
âˆ´ âˆ ACE = âˆ AEC
\ AC = AE ( by isosceles Î” theorem)
Now in Î” BAD, CE  DA
AE = DC (BPT)
AB BD
But AC = AE (proved above)
\ AC = DC
AB BD or
(proved).
Question2
State and prove the converse of angle bisector theorem.
Solution:
Given: ABC is a Î” ; AD divides BC in the ratio of the sides containing the angles âˆ A to meet BC at D.
i.e.
To prove: AD bisects âˆ A.
Construction: Draw CE  DA to meet BA produced at E.
Proof: In Î” ABC, CE  DA cut by AE.
\ âˆ BAD = âˆ AEC (corresponding angle) (i)
Similarly CE  DA cut by AC
\ âˆ DAC = âˆ ACE (alternate angles) (ii)
In DBEC; CE  AD
\ AB = BD (BPT)
AE DC
But AB = BD (given)
AC DC
\ AB = AB
AE AC
\ AE = AC
â‡’ âˆ AEC = âˆ ACE (isosceles property) (iii)
According to equation (i), (ii) and (iii) âˆ BAD = âˆ DAC
â‡’ AD bisects âˆ A.
Solution:
Given: ABC is a Î” ; AD divides BC in the ratio of the sides containing the angles âˆ A to meet BC at D.
i.e.
To prove: AD bisects âˆ A.
Construction: Draw CE  DA to meet BA produced at E.
Proof: In Î” ABC, CE  DA cut by AE.
\ âˆ BAD = âˆ AEC (corresponding angle) (i)
Similarly CE  DA cut by AC
\ âˆ DAC = âˆ ACE (alternate angles) (ii)
In DBEC; CE  AD
\ AB = BD (BPT)
AE DC
But AB = BD (given)
AC DC
\ AB = AB
AE AC
\ AE = AC
â‡’ âˆ AEC = âˆ ACE (isosceles property) (iii)
According to equation (i), (ii) and (iii) âˆ BAD = âˆ DAC
â‡’ AD bisects âˆ A.
Question3
If a parallelogram has all its sides equal and one of its diagonal is equal to a side, show that its diagonals are in the ratio : 1.
Solution:
Given: ABCD is a parallelogram, where AC and BD are the diagonals meeting at O. AB = BC = AC.
To Prove: BD : AC :: : 1
Proof: In Î” ABC, AB = BC = CA (given).
= a (say)
Hence ABC is an equilateral triangle. (Definition of equilateral triangle)
AC and BD are the diagonals of parallelogram ABCD,
â‡’ AC = BD (Diagonals of a parallelogram bisect each other)
or AO = OC.
i.e BO is the median of the equilateral ABC.
Hence BO =a
âˆ´ BD = a
â‡’ BD : AC :: a : a
â‡’ BD : AC :: : 1.
Solution:
Given: ABCD is a parallelogram, where AC and BD are the diagonals meeting at O. AB = BC = AC.
To Prove: BD : AC :: : 1
Proof: In Î” ABC, AB = BC = CA (given).
= a (say)
Hence ABC is an equilateral triangle. (Definition of equilateral triangle)
AC and BD are the diagonals of parallelogram ABCD,
â‡’ AC = BD (Diagonals of a parallelogram bisect each other)
or AO = OC.
i.e BO is the median of the equilateral ABC.
Hence BO =a
âˆ´ BD = a
â‡’ BD : AC :: a : a
â‡’ BD : AC :: : 1.
Question4
In Î” ABC, P, Q are points on AB and AC respectively and PQ  BC. Prove that the median AD bisects PQ.
Solution:
Given: ABC is a triangle, PQ  BC; AD is the median which cuts PQ at R.
To prove: AD bisects PQ at R.
Proof: In Î” ABD; PR  BD
AP = AR (BPT)
PB RD
In Î” ACD, RQ  DC
âˆ´ AR = AQ (BPT)
RD QC
In Î”APR and Î”ABD,
âˆ APR = âˆ ABD (corresponding angles.)
âˆ ARP = âˆ ADB (corresponding angles.)
âˆ´ Î” APR is similar to Î” ABD (AA similarity)
âˆ´ AP = AR = PR (corresponding sides of similar triangles are proportional)(i)
AB AD BD
Similarly Î” ARQ is similar to Î” ADC
\ AQ = AR = RQ (ii)
AC AD DC
According to equation (i) and (ii),
AR = PR = RQ
AD BD DC
but BD = DC (given)
âˆ´ PR = RQ
or AD bisects PQ at R (proved).
Solution:
Given: ABC is a triangle, PQ  BC; AD is the median which cuts PQ at R.
To prove: AD bisects PQ at R.
Proof: In Î” ABD; PR  BD
AP = AR (BPT)
PB RD
In Î” ACD, RQ  DC
âˆ´ AR = AQ (BPT)
RD QC
In Î”APR and Î”ABD,
âˆ APR = âˆ ABD (corresponding angles.)
âˆ ARP = âˆ ADB (corresponding angles.)
âˆ´ Î” APR is similar to Î” ABD (AA similarity)
âˆ´ AP = AR = PR (corresponding sides of similar triangles are proportional)(i)
AB AD BD
Similarly Î” ARQ is similar to Î” ADC
\ AQ = AR = RQ (ii)
AC AD DC
According to equation (i) and (ii),
AR = PR = RQ
AD BD DC
but BD = DC (given)
âˆ´ PR = RQ
or AD bisects PQ at R (proved).
Question5
Prove that the line joining the midpoints of the sides of the triangle form four triangles, each of which is similar to the original triangle.
Solution:
Given: In Î”ABC, D, E, F are the midpoints of AB, BC and AC respectively.
To prove: Î”ABC ~ Î”DEF
Î”ABC ~ Î”ADF
Î”ABC ~ Î”BDE
Î”ABC ~ Î”EFC
Proof: In Î”ABC, D and F are mid points of AB and AC respectively.
âˆ´ DF  BC (midpoint theorem)
In Î” ABC and Î” ADF
âˆ A is common; âˆ ADF = âˆ ABC (corresponding angles)
Î”ABC ~ Î” DF (AA similarity) (1)
Similarly we can prove Î”ABC ~ Î”BDE (AA similarity)(2)
Î”ABC ~ Î”EFC (AA similarity)(3)
In Î”ABC and Î”DEF;
since D,E, F are the midpoints of AB, BC and AC respectively,
DF= (1/2) Ã— BC; DE = (1/2) Ã— AC; EF = (1/2) Ã— AB; (midpoint theorem )
âˆ´ AB = BC = CA = 2
EF DF DE
âˆ´ Î” ABC ~ Î” EFD (SSS similarity )(4)
From (1), (2), (3) and (4)
Î” ABC ~ Î” DEF
Î” ABC ~ Î” ADF
Î” ABC ~ Î” BDE
Î” ABC ~ Î” EFC.
Solution:
Given: In Î”ABC, D, E, F are the midpoints of AB, BC and AC respectively.
To prove: Î”ABC ~ Î”DEF
Î”ABC ~ Î”ADF
Î”ABC ~ Î”BDE
Î”ABC ~ Î”EFC
Proof: In Î”ABC, D and F are mid points of AB and AC respectively.
âˆ´ DF  BC (midpoint theorem)
In Î” ABC and Î” ADF
âˆ A is common; âˆ ADF = âˆ ABC (corresponding angles)
Î”ABC ~ Î” DF (AA similarity) (1)
Similarly we can prove Î”ABC ~ Î”BDE (AA similarity)(2)
Î”ABC ~ Î”EFC (AA similarity)(3)
In Î”ABC and Î”DEF;
since D,E, F are the midpoints of AB, BC and AC respectively,
DF= (1/2) Ã— BC; DE = (1/2) Ã— AC; EF = (1/2) Ã— AB; (midpoint theorem )
âˆ´ AB = BC = CA = 2
EF DF DE
âˆ´ Î” ABC ~ Î” EFD (SSS similarity )(4)
From (1), (2), (3) and (4)
Î” ABC ~ Î” DEF
Î” ABC ~ Î” ADF
Î” ABC ~ Î” BDE
Î” ABC ~ Î” EFC.
Question6
In triangle ABC, DEBC and AD : DB = 2 : 3. Determine the ratio of the area triangle ADE to the area triangle ABC.
Solution:
âˆ´ ==
Solution:
âˆ´ ==
Question7
One angle of a triangle is equal to one angle of another triangle and the bisectors of these equal angles divide the opposite sides in the same ratio. Prove that the triangles are similar.
Solution:
Given: Î”ABC and Î”PQR
âˆ A = âˆ P
AD and PS bisects âˆ A and âˆ P respectively.
BD = QS
DC SR
To prove: Î”ABC ~ Î”PQR
Proof: In Î”ABC and Î”PQR
AD bisects âˆ A
âˆ´ AB = BD (Angle bisector theorem) (1)
AC DC
Similarly in Î”PQR,
PQ = QS (Angle bisector theorem) (2)
PR SR
But BD = QS (given)
DC SR
âˆ´ According to equation (1) and (2)
AB = PQ Ãž AB = AC
AC PR PQ PR
âˆ A = âˆ P (given)
âˆ´ Î”ABC ~ Î”PQR (SAS similarity).
Solution:
Given: Î”ABC and Î”PQR
âˆ A = âˆ P
AD and PS bisects âˆ A and âˆ P respectively.
BD = QS
DC SR
To prove: Î”ABC ~ Î”PQR
Proof: In Î”ABC and Î”PQR
AD bisects âˆ A
âˆ´ AB = BD (Angle bisector theorem) (1)
AC DC
Similarly in Î”PQR,
PQ = QS (Angle bisector theorem) (2)
PR SR
But BD = QS (given)
DC SR
âˆ´ According to equation (1) and (2)
AB = PQ Ãž AB = AC
AC PR PQ PR
âˆ A = âˆ P (given)
âˆ´ Î”ABC ~ Î”PQR (SAS similarity).
Question8
The bisector of interior angle A of a triangle AB meets BC in D and the bisector of exterior angle A meets BC produced in E. Prove that
Solution:
Given: Î”ABC, AD bisects interior âˆ A and AE bisects exterior âˆ A meeting BC at D and BC produced at E.
To prove:
Proof: In Î”ABC, AD bisects interior âˆ A
âˆ´ (Angle Bisector theorem)â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)
Similarly in DABC, AE bisects exterior âˆ A
âˆ´ â€¦â€¦â€¦â€¦â€¦â€¦..(2)
From equation (1) and (2),
Hence Proved.
Solution:
Given: Î”ABC, AD bisects interior âˆ A and AE bisects exterior âˆ A meeting BC at D and BC produced at E.
To prove:
Proof: In Î”ABC, AD bisects interior âˆ A
âˆ´ (Angle Bisector theorem)â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)
Similarly in DABC, AE bisects exterior âˆ A
âˆ´ â€¦â€¦â€¦â€¦â€¦â€¦..(2)
From equation (1) and (2),
Hence Proved.
Question9
In a triangle ABC, XY  AC divides the triangle into two parts equal in areas. Determine .
Solution:
Given: ABC is a triangle with XY  AC divides the triangle into two parts equal in areas.
To find:
Proof:
ar Î”BXY = ar trap. XYCA (Given) âˆ´ ar Î”BXY = ar Î”ABC
In Î”BXY andBAC,
âˆ BXY = âˆ BAC (Corresponding angles)
âˆ BYX = âˆ BCA (Corresponding angles)
Î”BXY âˆ¼ Î”BAC (AA similarity)
âˆ´ = (Areas of similar triangle)
âˆ´ =
âˆ´ =
âˆ´ AB â€“ BX = âˆš 2 BX â€“ BX
âˆ´ AX = (âˆš 2 â€“ 1)BX
= =.
Solution:
Given: ABC is a triangle with XY  AC divides the triangle into two parts equal in areas.
To find:
Proof:
ar Î”BXY = ar trap. XYCA (Given) âˆ´ ar Î”BXY = ar Î”ABC
In Î”BXY andBAC,
âˆ BXY = âˆ BAC (Corresponding angles)
âˆ BYX = âˆ BCA (Corresponding angles)
Î”BXY âˆ¼ Î”BAC (AA similarity)
âˆ´ = (Areas of similar triangle)
âˆ´ =
âˆ´ =
âˆ´ AB â€“ BX = âˆš 2 BX â€“ BX
âˆ´ AX = (âˆš 2 â€“ 1)BX
= =.
Question10
If two sides and a median bisecting the third side of a Î” are respectively proportional to the corresponding sides and the median of another triangle, then prove that the two triangles are similar.
Solution:
Given: Î”ABC and Î”PQR where BD and QS are the medians and AB = BC = BD
PQ QR QS
To prove: Î”ABC ~ Î”PQR
Construction: Produce BD and QS to E and T respectively such that BD = DE and QS = ST. CE and TR are joined.
Proof: In Î”ADB and DCDE,
AD = DC (given)
âˆ ADB = âˆ CDE (Vertically opposite angles)
BD = DE.
âˆ´ Î”ADB â‰… Î”CDE (SAS â‰… axiom)
Hence AB = CE and âˆ ABD = âˆ DEC.
Similarly Î”PQS â‰… Î” RST,
hence PQ = TR and âˆ PQS = âˆ STR.
Consider Î” EBC and Î” TQR,
BD = 2 BD = BE (from given and construction)(1)
QS 2QS QT
AB = CE and PQ = RT (proved),
AB = CE (2)
PQ RT
AB = BC = BD (Given) (3)
PQ QR QS
From (1),(2) and (3),
BE = CE = BC
QT RT QR
âˆ´ Î”EBC ~ Î”TQR(SSS similarity axiom).
â‡’ âˆ DBC = âˆ SQR and âˆ DEC = âˆ STR (4) (corresponding angles of similar triangles are proportional)
But âˆ ABD = âˆ DEC and âˆ PQS = âˆ STR (proved)(5)
âˆ´ âˆ ABD = âˆ PQS (from (4) and (5)) (6)
From (5) and (6),
âˆ ABC = âˆ PQR I
In Î” ABC and Î” PQR,
AB = BC (given)
PQ QR
And âˆ ABC = âˆ PQR (from I)
âˆ´ Î”ABC ~ Î”PQR (SAS Similarity).
Solution:
Given: Î”ABC and Î”PQR where BD and QS are the medians and AB = BC = BD
PQ QR QS
To prove: Î”ABC ~ Î”PQR
Construction: Produce BD and QS to E and T respectively such that BD = DE and QS = ST. CE and TR are joined.
Proof: In Î”ADB and DCDE,
AD = DC (given)
âˆ ADB = âˆ CDE (Vertically opposite angles)
BD = DE.
âˆ´ Î”ADB â‰… Î”CDE (SAS â‰… axiom)
Hence AB = CE and âˆ ABD = âˆ DEC.
Similarly Î”PQS â‰… Î” RST,
hence PQ = TR and âˆ PQS = âˆ STR.
Consider Î” EBC and Î” TQR,
BD = 2 BD = BE (from given and construction)(1)
QS 2QS QT
AB = CE and PQ = RT (proved),
AB = CE (2)
PQ RT
AB = BC = BD (Given) (3)
PQ QR QS
From (1),(2) and (3),
BE = CE = BC
QT RT QR
âˆ´ Î”EBC ~ Î”TQR(SSS similarity axiom).
â‡’ âˆ DBC = âˆ SQR and âˆ DEC = âˆ STR (4) (corresponding angles of similar triangles are proportional)
But âˆ ABD = âˆ DEC and âˆ PQS = âˆ STR (proved)(5)
âˆ´ âˆ ABD = âˆ PQS (from (4) and (5)) (6)
From (5) and (6),
âˆ ABC = âˆ PQR I
In Î” ABC and Î” PQR,
AB = BC (given)
PQ QR
And âˆ ABC = âˆ PQR (from I)
âˆ´ Î”ABC ~ Î”PQR (SAS Similarity).
Question11
Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same sides of BC. If AC and DB intersect at P, prove that AP Ã— PC = BP Ã— PD.
Solution:
Given: Two right triangles ABC and BDC on the same hypotenuse BC. AC and BD intersect at P.
To prove: AP Ã— PC = BP Ã— PD
Proof: In Î”ABP and Î”DCP
âˆ A = âˆ D (= 90Â°) (given)
âˆ APB = âˆ DPC (vertically opposite angles)
âˆ´ Î”ABP ~ Î”DCP (AA similarity axiom)
âˆ´ AB = BP = AP (corresponding sides of similar Î” s are proportional)â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(1)
DC CP DP
From (1) BP = AP
CP DP
By cross multiplication,
BP Ã— DP = AP Ã— PC (proved).
Solution:
Given: Two right triangles ABC and BDC on the same hypotenuse BC. AC and BD intersect at P.
To prove: AP Ã— PC = BP Ã— PD
Proof: In Î”ABP and Î”DCP
âˆ A = âˆ D (= 90Â°) (given)
âˆ APB = âˆ DPC (vertically opposite angles)
âˆ´ Î”ABP ~ Î”DCP (AA similarity axiom)
âˆ´ AB = BP = AP (corresponding sides of similar Î” s are proportional)â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(1)
DC CP DP
From (1) BP = AP
CP DP
By cross multiplication,
BP Ã— DP = AP Ã— PC (proved).
Question12
If ABC is an equilateral triangle of side 2a prove that the altitude AD = 3a and 3AB^{2} = 4AD^{2}.
Solution:
Given: Î” ABC is an equilateral triangle of side 2a. AD is the altitude of the triangle.
To Prove: AD = aâˆš 3 and 3AB^{2} = 4AD^{2}^{ }
Proof:
In rt. Î” ADC,
AD^{2} = AC^{2}  DC^{2}
= (2a)^{2}  a^{2}
= 4a^{2}  a^{2}
= 3a^{2}
âˆ´^{ } AD = aâˆš3
3AB^{2 }= 3(2a)^{2 }
= 3(2a)^{2}
= 3 Â´ 4a^{2}
= 4(aâˆš 3)2
= 4AD^{2}
âˆ´ 3AB^{2} = 4AD^{2}.
Solution:
Given: Î” ABC is an equilateral triangle of side 2a. AD is the altitude of the triangle.
To Prove: AD = aâˆš 3 and 3AB^{2} = 4AD^{2}^{ }
Proof:
In rt. Î” ADC,
AD^{2} = AC^{2}  DC^{2}
= (2a)^{2}  a^{2}
= 4a^{2}  a^{2}
= 3a^{2}
âˆ´^{ } AD = aâˆš3
3AB^{2 }= 3(2a)^{2 }
= 3(2a)^{2}
= 3 Â´ 4a^{2}
= 4(aâˆš 3)2
= 4AD^{2}
âˆ´ 3AB^{2} = 4AD^{2}.
Question13
Two isosceles Î”s have equal vertical angles and their areas are in the ratio 9 : 16. Find the ratio of their corresponding heights (altitudes).
Solution:
Given: Î”ABC and Î”PQR are isosceles and âˆ A = âˆ P. AD, PS are the altitudes and .
To find: AD
PS
Proof: In Î” ABC, âˆ B = âˆ C (isosceles Î” property)
Solution:
Given: Î”ABC and Î”PQR are isosceles and âˆ A = âˆ P. AD, PS are the altitudes and .
To find: AD
PS
Proof: In Î” ABC, âˆ B = âˆ C (isosceles Î” property)
Similarly in Î” PQR, âˆ Q = âˆ R.
âˆ A = âˆ P (given)
âˆ´ âˆ B = âˆ C =
Since âˆ A = âˆ P
âˆ B = âˆ C = âˆ Q = âˆ R
âˆ´ Î” ABC ~ Î” PQR (AA)
If 2 triangles are similar then the ratio of areas will be equal to the square of the corresponding sides,
â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)
In Î” ABD, Î” PQS
âˆ D = âˆ S (= 90Â° )
âˆ B = âˆ Q (given)
âˆ´ Î” ABD ~ Î” PQS (AA)
âˆ´ AB = AD â€¦â€¦â€¦â€¦â€¦â€¦.(2)
PQ PS
According equation (1)
âˆ´
â‡’ AD = 3
PS 4
.^{.}. The ratio of their corresponding heights is 3 : 4.
Question14
The diagonals of a quadrilateral ABCD intersect each other at the point O such that . Show that ABCD is a trapezium.
Solution:
Given: ABCD is a trapezium with AB  CD and the diagonals AC and BD intersect at â€˜Oâ€™.
To prove: =
Proof:
In the figure consider the triangle OAB and OCD
âˆ DOC = âˆ Î‘OB (Vertically opposite angles are equal)
since AB  DC,
âˆ DCO = âˆ OAB (Alternate angles are equal)
âˆ´ By AA corollary of similar triangles.
âˆ´ Î” OAB âˆ¼ Î” OCB When the two triangle are similar, the side are proportionally.
â‡’ =
Hence proved.
Solution:
Given: ABCD is a trapezium with AB  CD and the diagonals AC and BD intersect at â€˜Oâ€™.
To prove: =
Proof:
In the figure consider the triangle OAB and OCD
âˆ DOC = âˆ Î‘OB (Vertically opposite angles are equal)
since AB  DC,
âˆ DCO = âˆ OAB (Alternate angles are equal)
âˆ´ By AA corollary of similar triangles.
âˆ´ Î” OAB âˆ¼ Î” OCB When the two triangle are similar, the side are proportionally.
â‡’ =
Hence proved.
Question15
P and Q are the points on the sides AB and AC respectively of a Î”ABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cms, QC = 6 cm, prove that BC = 3PQ.
Solution:
Given: Î” ABC, PQ are points on AB and AC such that AP = 2 cm, BP = 4 cm, AQ = 3 cm, QC = 6 cm
To prove: BC = 3PQ
Proof: In Î” ABC,
As AP = AQ
PB QC
According to converse of BPT, PQ  BC
In Î” APQ and Î” ABC
âˆ´ âˆ APQ = âˆ ABC (Corresponding angles)
âˆ A is Common
âˆ´ Î” APQ ~ Î” ABC (AAS similarity)
(corresponding sides of similar Î” s are proportional)
But
âˆ´ PQ = 2 = 1
BC 6 3
âˆ´ 3PQ = BC (Proved).
Solution:
Given: Î” ABC, PQ are points on AB and AC such that AP = 2 cm, BP = 4 cm, AQ = 3 cm, QC = 6 cm
To prove: BC = 3PQ
Proof: In Î” ABC,
As AP = AQ
PB QC
According to converse of BPT, PQ  BC
In Î” APQ and Î” ABC
âˆ´ âˆ APQ = âˆ ABC (Corresponding angles)
âˆ A is Common
âˆ´ Î” APQ ~ Î” ABC (AAS similarity)
(corresponding sides of similar Î” s are proportional)
But
âˆ´ PQ = 2 = 1
BC 6 3
âˆ´ 3PQ = BC (Proved).
Question16
Through the midpoint of M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that EL = 2BL.
Solution:
Given: ABCD is a parallelogram, M is the midpoint of CD. BM intersects AC at L and AD produced at E.
To prove: EL = 2BL
Proof: In Î” BMC and Î” EDM
âˆ DME = âˆ BMC (Vetically opposite angles)
DM = MC (given)
âˆ DEM = âˆ MBC (alternate angles)
âˆ´ Î” BMC â‰… Î” EDM (ASA congruence)
âˆ´ DE = BC (c.p.c.t)
But BC = AD (opposite sides of parallelogram ABCD)
âˆ´ AD = DE â‡’ AE = 2AD = 2BC
In Î” AEL and Î” CBL
âˆ ALE = âˆ BLC (Vertically opposite angles)
âˆ AEL = âˆ LBC (alternate angles)
âˆ´ Î” AEL ~ Î” CBL (AA similarity axiom)
âˆ´ EL = 2 BL
Solution:
Given: ABCD is a parallelogram, M is the midpoint of CD. BM intersects AC at L and AD produced at E.
To prove: EL = 2BL
Proof: In Î” BMC and Î” EDM
âˆ DME = âˆ BMC (Vetically opposite angles)
DM = MC (given)
âˆ DEM = âˆ MBC (alternate angles)
âˆ´ Î” BMC â‰… Î” EDM (ASA congruence)
âˆ´ DE = BC (c.p.c.t)
But BC = AD (opposite sides of parallelogram ABCD)
âˆ´ AD = DE â‡’ AE = 2AD = 2BC
In Î” AEL and Î” CBL
âˆ ALE = âˆ BLC (Vertically opposite angles)
âˆ AEL = âˆ LBC (alternate angles)
âˆ´ Î” AEL ~ Î” CBL (AA similarity axiom)
âˆ´ EL = 2 BL
Question17
The side BC of a triangle ABC is bisected at D; O is any point in AD. BO, CO produced meet AC, AB in E,F respectively, and AD is produced to X so that D is the mid point of OX. Prove that AO : AX = AF : AB and show that EF is parallel to BC.
Solution:
Given : The side BC of a triangle ABC is bisected at D; O is any point in AD. BO , CO produced meet AC, AB in E , F respectively, and AD is produced to X so that D is the mid point of OX.
To Prove : AO : AX = AF : AB and show that EF is parallel to BC.
Construction: Join BX and CX.
Proof: In quadrilateral BOCX, BD = DC and DO = DX (given)
âˆ´ BOCX is a parallelogram (When the diagonals of a quadrilateral bisect each other, then the quad. is a parallelogram)
âˆ´ BX  CO (Definition of a parallelogram)
or BX  FO.
In Î” ABX, BX  FO(proved).
âˆ´ AO : AX = AF : AB (using B.P.T) (i)
Similarly, AO : AX = AE : AC (ii)
From (i) and (ii),
AF : AB = AE : AC
By corollary to B.P.T, EF is parallel to BC.
Solution:
Given : The side BC of a triangle ABC is bisected at D; O is any point in AD. BO , CO produced meet AC, AB in E , F respectively, and AD is produced to X so that D is the mid point of OX.
To Prove : AO : AX = AF : AB and show that EF is parallel to BC.
Construction: Join BX and CX.
Proof: In quadrilateral BOCX, BD = DC and DO = DX (given)
âˆ´ BOCX is a parallelogram (When the diagonals of a quadrilateral bisect each other, then the quad. is a parallelogram)
âˆ´ BX  CO (Definition of a parallelogram)
or BX  FO.
In Î” ABX, BX  FO(proved).
âˆ´ AO : AX = AF : AB (using B.P.T) (i)
Similarly, AO : AX = AE : AC (ii)
From (i) and (ii),
AF : AB = AE : AC
By corollary to B.P.T, EF is parallel to BC.
Question18
ABC is a triangle in which âˆ BAC = 90Â° and DEFG is a square, prove that DE^{2} = BD Ã— EC.
Solution:
Given: ABC is a triangle in which âˆ BAC = 90Â° and DEFG is a square.
To prove: DE^{2} = BD Ã— EC.
Proof: In Î” AGF and Î” DBG,
âˆ AGF = âˆ GBD (corresponding angles)
âˆ GAF = âˆ BDG (each = 90Â° )
âˆ´ Î”AGF âˆ¼ Î”DBG. (i)
Similarly, Î”AFG âˆ¼ Î”ECF (AA Similarity)(ii)
From (i) and (ii), Î”DBG âˆ¼ Î”ECF.
EF Ã— DG = BD Ã— EC. (iii)
Also DEFG is a square â‡’ DE = EF = FG = DG (iv)
From (iii) and (iv), DE^{2} = BD Ã— EC.
Solution:
Given: ABC is a triangle in which âˆ BAC = 90Â° and DEFG is a square.
To prove: DE^{2} = BD Ã— EC.
Proof: In Î” AGF and Î” DBG,
âˆ AGF = âˆ GBD (corresponding angles)
âˆ GAF = âˆ BDG (each = 90Â° )
âˆ´ Î”AGF âˆ¼ Î”DBG. (i)
Similarly, Î”AFG âˆ¼ Î”ECF (AA Similarity)(ii)
From (i) and (ii), Î”DBG âˆ¼ Î”ECF.
EF Ã— DG = BD Ã— EC. (iii)
Also DEFG is a square â‡’ DE = EF = FG = DG (iv)
From (iii) and (iv), DE^{2} = BD Ã— EC.
Question19
In fig. DE  BC. If AD = x, DB = x â€“ 2, AE = x + 2 and EC = x â€“ 1, find the value x.
Solution:
Given: ABC is a triangle, DE  BC, AD = x, DB = x â€“ 2, AE = x + 2 and EC = x â€“ 1.
To find: x
Solution:
Given: ABC is a triangle, DE  BC, AD = x, DB = x â€“ 2, AE = x + 2 and EC = x â€“ 1.
To find: x
In Î” ABC, we have DE  BC Therefore [By Thaleâ€™s theorem] AD Ã— EC = AE Ã— DB x(x  1) = (x  2)(x + 2) x^{2} â€“ x = x^{2} â€“ 4 x = 4 

Question20
In a Î” ABC, D and E are points on the sides AB and AC respectively such that DE  BC. If AD = 4 cm, AE = 8 cm, DB = x â€“ 4
and EC = 3x â€“ 19, find x.
Solution:
and EC = 3x â€“ 19, find x.
Solution:
Given: In Î” ABC, D and E are points on the sides AB and AC respectively such that DE  BC. AD = 4 cm, AE = 8 cm, DB = x  4 and EC = 3x  19. To find: x. In Î” ABC, we have DE  BC Therefore [By Thaleâ€™s theorem] 4(3x  19) = 8(x  4) 12x â€“ 76 = 8x â€“ 32 4x = 44 x = 11 
Question21
In a Î” ABC, AD is the bisector of âˆ A, meeting side BC at D. If AC = 4.2 cm, DC = 6 cm, BC = 10 cm, find AB.
Solution:
Solution:
Given: In a Î” ABC, AD is the bisector of âˆ A, meeting side BC at D. AC = 4.2 cm, DC = 6 cm and BC = 10 cm. To find: AB.

Question22
In Î” ABC, âˆ B = 90Â° and D is the midpoint of BC. Prove that AC^{2} = AD^{2} + 3CD^{2}.
Solution:
Given: In Î” ABC, âˆ B = 90Â° and D is the midpoint of BC.
To Prove: AC^{2} = AD^{2} + 3CD^{2}
Proof:
In Î” ABD,
AD^{2} = AB^{2} + BD^{2}
AB^{2 }= AD^{2}  BD^{2 }â€¦â€¦â€¦â€¦â€¦..(i)
In Î” ABC,
AC^{2} = AB^{2} + BC^{2}
AB^{2 }= AC^{2}  BC^{2 }â€¦â€¦â€¦â€¦â€¦..(ii)
Equating (i) and (ii)
AD^{2}  BD^{2 }= AC^{2}  BC^{2}
AD^{2}  BD^{2 }= AC^{2} â€“ (BD + DC)^{2}
AD^{2}  BD^{2 }= AC^{2} â€“ BD^{2}  DC^{2} â€“ 2BD Ã— DC
AD^{2} = AC^{2}  DC^{2} â€“ 2DC^{2 }(DC = BD)
AD^{2} = AC^{2}  3DC^{2}
Solution:
Given: In Î” ABC, âˆ B = 90Â° and D is the midpoint of BC.
To Prove: AC^{2} = AD^{2} + 3CD^{2}
Proof:
In Î” ABD,
AD^{2} = AB^{2} + BD^{2}
AB^{2 }= AD^{2}  BD^{2 }â€¦â€¦â€¦â€¦â€¦..(i)
In Î” ABC,
AC^{2} = AB^{2} + BC^{2}
AB^{2 }= AC^{2}  BC^{2 }â€¦â€¦â€¦â€¦â€¦..(ii)
Equating (i) and (ii)
AD^{2}  BD^{2 }= AC^{2}  BC^{2}
AD^{2}  BD^{2 }= AC^{2} â€“ (BD + DC)^{2}
AD^{2}  BD^{2 }= AC^{2} â€“ BD^{2}  DC^{2} â€“ 2BD Ã— DC
AD^{2} = AC^{2}  DC^{2} â€“ 2DC^{2 }(DC = BD)
AD^{2} = AC^{2}  3DC^{2}
Question23
ABC is a triangle in which AB = AC and D is a point on the side AC such that BC^{2} = AC Ã— CD. Prove that BD = BC.
Solution:
Given: A Î” ABC in which AB = AC. D is a point on AC such that BC^{2} = AC Ã— CD.
To prove: BD = BC
Proof: Since BC^{2} = AC Ã— CD
Therefore BC Ã— BC = AC Ã— CD
AC/BC = BC/CD â€¦â€¦â€¦â€¦â€¦(i)
Also âˆ ACB = âˆ BCD
Since Î”ABC âˆ¼ Î”BDC [By SAS Axiom of similar triangles]
AB/AC = BD/BC â€¦â€¦â€¦â€¦â€¦..(ii)
But AB = AC (Given) â€¦â€¦â€¦â€¦â€¦(iii)
From (i), (ii) and (iii) we get
BD = BC.
Solution:
Given: A Î” ABC in which AB = AC. D is a point on AC such that BC^{2} = AC Ã— CD.
To prove: BD = BC
Proof: Since BC^{2} = AC Ã— CD
Therefore BC Ã— BC = AC Ã— CD
AC/BC = BC/CD â€¦â€¦â€¦â€¦â€¦(i)
Also âˆ ACB = âˆ BCD
Since Î”ABC âˆ¼ Î”BDC [By SAS Axiom of similar triangles]
AB/AC = BD/BC â€¦â€¦â€¦â€¦â€¦..(ii)
But AB = AC (Given) â€¦â€¦â€¦â€¦â€¦(iii)
From (i), (ii) and (iii) we get
BD = BC.
Question24
The diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Prove that
DF Ã— EF = FB Ã— FA.
Solution:
DF Ã— EF = FB Ã— FA.
Solution:
Given: The diagonal BD of parallelogram ABCD intersects the segment AE at F, where E is any point on BC. To prove: DF Ã— EF = FB Ã— FA Proof: In triangles AFD and BFE, âˆ FAD = âˆ FEB (Alternate angles) âˆ AFD = âˆ BFE (Vertically opposite angles) Therefore Î”ADF âˆ¼ Î”BFE (AA similarity) Hence DF Ã— EF = FB Ã— FA 
Question25
ABC is a triangle, rightangled at C and AC = âˆš3 BC. Prove that âˆ ABC = 60Â°.
Solution:
Given: Î”ABC is right angled at C and AC = âˆš3BC.
To prove: âˆ ABC = 60Â°.
Proof:
Let D be the midpoint of AB. Join CD.
Now, AB^{2} = BC^{2} + AC^{2} = BC^{2} + (BC)^{2} = 4BC^{2}
Therefore AB = 2BC.
Now, BD = AB = (2BC) = BC.
But, D being the midpoint of hypotenuse AB, it is equidistant from all the three vertices.
Therefore CD = BD = DA or CD = AB = BC.
Thus, BC = BD = CD,
i.e., Î”BCD is a equilateral triangle.
Hence, âˆ ABC = 60Â°.
Solution:
Given: Î”ABC is right angled at C and AC = âˆš3BC.
To prove: âˆ ABC = 60Â°.
Proof:
Let D be the midpoint of AB. Join CD.
Now, AB^{2} = BC^{2} + AC^{2} = BC^{2} + (BC)^{2} = 4BC^{2}
Therefore AB = 2BC.
Now, BD = AB = (2BC) = BC.
But, D being the midpoint of hypotenuse AB, it is equidistant from all the three vertices.
Therefore CD = BD = DA or CD = AB = BC.
Thus, BC = BD = CD,
i.e., Î”BCD is a equilateral triangle.
Hence, âˆ ABC = 60Â°.
Question26
ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.
Solution:
ble border="0" cellpadding="1" cellspacing="0" width="582">
Given: ABCD is a quadrilateral in which AD = BC. P, Q, R and S be the midpoints of AB, AC, CD and BD respectively.
To prove: PQRS is a rhombus.
Proof: By a theorem, line segment joining the mid points of any two sides of a triangle is parallel to the third side and equal to half of it.
In Î” ABC, PQBC and PQ = BC = DA
In Î” CDA, RQDA and RQ = DA
In Î” BDA, SPDA and SP = DA
In Î” CDB, SRBC and SR = BC = DA
Therefore SP  RQ, PQ  SR and PQ = RQ = SP = SR.
Hence PQRS is a rhombus.
Solution:
ble border="0" cellpadding="1" cellspacing="0" width="582">
Given: ABCD is a quadrilateral in which AD = BC. P, Q, R and S be the midpoints of AB, AC, CD and BD respectively.
To prove: PQRS is a rhombus.
Proof: By a theorem, line segment joining the mid points of any two sides of a triangle is parallel to the third side and equal to half of it.
In Î” ABC, PQBC and PQ = BC = DA
In Î” CDA, RQDA and RQ = DA
In Î” BDA, SPDA and SP = DA
In Î” CDB, SRBC and SR = BC = DA
Therefore SP  RQ, PQ  SR and PQ = RQ = SP = SR.
Hence PQRS is a rhombus.
Question27
Let ABC be a triangle, rightangled at C. If D is the midpoint of BC, prove that AB^{2 }= 4AD^{2} â€“^{ }3AC^{2}.
Solution:
Given: ABC be a triangle, rightangled at C and D is the midpoint of BC.
To Prove: AB^{2 }= 4AD^{2} â€“^{ }3AC^{2}.
Proof:
From right triangle ACB, we have,
AB^{2} = AC^{2} + BC^{2}
= AC^{2} + (2CD)^{2} = AC^{2} + 4CD^{2 }[Since BC = 2CD]
= AC^{2} + 4(AD^{2}  AC^{2}) [From right Î” ACD]
= 4AD^{2} â€“^{ }3AC^{2}.
Solution:
Given: ABC be a triangle, rightangled at C and D is the midpoint of BC.
To Prove: AB^{2 }= 4AD^{2} â€“^{ }3AC^{2}.
Proof:
From right triangle ACB, we have,
AB^{2} = AC^{2} + BC^{2}
= AC^{2} + (2CD)^{2} = AC^{2} + 4CD^{2 }[Since BC = 2CD]
= AC^{2} + 4(AD^{2}  AC^{2}) [From right Î” ACD]
= 4AD^{2} â€“^{ }3AC^{2}.
Question28
In the given figure, points D and E trisect BC and âˆ B = 90Â°. Prove that 8AE^{2} = 3AC^{2} + 5AD^{2}.
Solution:
Given: In Î” ABC, points D and E trisect on BC and âˆ B = 90Â°.
To Prove: 8AE^{2} = 3AC^{2} + 5AD^{2}.
Proof: Let ABC be the triangle in which h âˆ B = 90Â°. Let the points D and E trisect BC.
Join AD and AE. Then,
AC^{2} = AB^{2} + BC^{2}
3AC^{2} = 3AB^{2} + 3BC^{2} â€¦â€¦â€¦..(i)
AD^{2} = AB^{2} + BD^{2}
5AD^{2} = 5AB^{2} + 5BD^{2} ..â€¦â€¦..(ii)
Therefore 3AC^{2} + 5AD^{2} = 8AB^{2} + 3BC^{2} + 5BD^{2 }
= 8AB^{2} + 3.+ 5
= 8AB^{2} +BE^{2}
= 8AB^{2} + 8BE^{2 }= 8(AB^{2} + BE^{2})
= 8AE^{2}.
Solution:
Given: In Î” ABC, points D and E trisect on BC and âˆ B = 90Â°.
To Prove: 8AE^{2} = 3AC^{2} + 5AD^{2}.
Proof: Let ABC be the triangle in which h âˆ B = 90Â°. Let the points D and E trisect BC.
Join AD and AE. Then,
AC^{2} = AB^{2} + BC^{2}
3AC^{2} = 3AB^{2} + 3BC^{2} â€¦â€¦â€¦..(i)
AD^{2} = AB^{2} + BD^{2}
5AD^{2} = 5AB^{2} + 5BD^{2} ..â€¦â€¦..(ii)
Therefore 3AC^{2} + 5AD^{2} = 8AB^{2} + 3BC^{2} + 5BD^{2 }
= 8AB^{2} + 3.+ 5
= 8AB^{2} +BE^{2}
= 8AB^{2} + 8BE^{2 }= 8(AB^{2} + BE^{2})
= 8AE^{2}.
Question29
In fig., ABC is a triangle in which AB = AC. D and E are points on the sides AB and AC respectively such that AD = AE. Show that the points B, C, E and D are concyclic.
Solution:
Given: In Î”ABC, AB = AC. D and E are points on the sides AB and AC respectively such that AD = AE.
To Prove: Points B, C, E and D are concyclic.
Proof: In order to prove that the points B, C, E and D are concyclic, it is sufficient to show that âˆ ABC + âˆ CED = 180Â° and
âˆ ACB + âˆ BDE = 180Â°.
In Î”ABC, we have
AB = AC and AD = AE
AB  AD = AC  AE
DB = EC
Thus, we have
AD = AE and DB = EC.
DE  BC [By the converse of Thale's Theorem]
âˆ ABC = âˆ ADE [Corresponding angles]
âˆ ABC + âˆ BDE = âˆ ADE + âˆ BDE [adding âˆ BDE on both sides]
âˆ ABC + âˆ BDE = 180Â°
âˆ ACB + âˆ BDE = 180Â° [Since AB = AC Therefore âˆ ABC = âˆ ACB]
Again DEBC
âˆ ACB = âˆ AED
âˆ ACB + âˆ CED = âˆ AED + âˆ CED [Adding âˆ CED on both sides]
âˆ ACB + âˆ CED = 180Â°
âˆ ABC + âˆ CED = 180Â°^{ }[Since âˆ ABC = âˆ ACB]
Therefore BDEC is a cyclic quadrilateral.
Hence, B, C, E and D are concyclic points.
Solution:
Given: In Î”ABC, AB = AC. D and E are points on the sides AB and AC respectively such that AD = AE.
To Prove: Points B, C, E and D are concyclic.
Proof: In order to prove that the points B, C, E and D are concyclic, it is sufficient to show that âˆ ABC + âˆ CED = 180Â° and
âˆ ACB + âˆ BDE = 180Â°.
In Î”ABC, we have
AB = AC and AD = AE
AB  AD = AC  AE
DB = EC
Thus, we have
AD = AE and DB = EC.
DE  BC [By the converse of Thale's Theorem]
âˆ ABC = âˆ ADE [Corresponding angles]
âˆ ABC + âˆ BDE = âˆ ADE + âˆ BDE [adding âˆ BDE on both sides]
âˆ ABC + âˆ BDE = 180Â°
âˆ ACB + âˆ BDE = 180Â° [Since AB = AC Therefore âˆ ABC = âˆ ACB]
Again DEBC
âˆ ACB = âˆ AED
âˆ ACB + âˆ CED = âˆ AED + âˆ CED [Adding âˆ CED on both sides]
âˆ ACB + âˆ CED = 180Â°
âˆ ABC + âˆ CED = 180Â°^{ }[Since âˆ ABC = âˆ ACB]
Therefore BDEC is a cyclic quadrilateral.
Hence, B, C, E and D are concyclic points.
Question30
In fig, âˆ B < 90Â° and segment AD ^ BC, show that b^{2} = h^{2} + a^{2} + x^{2}  2ax
Solution:
Given: In Î”ABC, âˆ B < 90Â° and segment AD âŠ¥ BC.
To prove: b^{2} = h^{2} + a^{2} + x^{2} â€“ 2ax
Proof:
b^{2} = h^{2} + (a  x)^{2}
b^{2} = h^{2} + a^{2} + x^{2}  2ax.
Given: In Î”ABC, âˆ B < 90Â° and segment AD âŠ¥ BC.
To prove: b^{2} = h^{2} + a^{2} + x^{2} â€“ 2ax
Proof:
b^{2} = h^{2} + (a  x)^{2}
b^{2} = h^{2} + a^{2} + x^{2}  2ax.
Question31
In the given figure, ABCD is a parallelogram P is a point on BC, such that BP : PC = 1 : 2. DP produced meets AB produced at Q. Given area of triangle CPQ = 20 m^{2}, calculate the area of triangle DCP.
Solution:
Given: ABCD is a parallelogram P is a point on BC, such that BP : PC = 1 : 2. DP produced meets AB produced at Q.
Area of triangle CPQ = 20 m^{2}.
âˆ BPQ = âˆ DPC (Vertically opposite angles)
âˆ BQP = âˆ PDC (alternate angles, BQ  DC, DQ meets them)
âˆ´ Î” BPQ âˆ¼ Î” CPD (AA similarity)
BP/CP = Â½ (Given)
âˆ´ Area D CPD = Area D BPQ
and area Î”CPQ = 20 cm^{2} (Given)
Area Î”BPQ = 10 cm^{2}
Area Î”CPD = 40 cm^{2}
(Proportional to bases BC and PC)
âˆ´ Area Î” DBC = 40 Ã— 3/2 = 60 cm^{2}.
Given: ABCD is a parallelogram P is a point on BC, such that BP : PC = 1 : 2. DP produced meets AB produced at Q.
Area of triangle CPQ = 20 m^{2}.
To Find: Area of triangle DCP. Construction: Join DB. 
âˆ BPQ = âˆ DPC (Vertically opposite angles)
âˆ BQP = âˆ PDC (alternate angles, BQ  DC, DQ meets them)
âˆ´ Î” BPQ âˆ¼ Î” CPD (AA similarity)
BP/CP = Â½ (Given)
âˆ´ Area D CPD = Area D BPQ
and area Î”CPQ = 20 cm^{2} (Given)
Area Î”BPQ = 10 cm^{2}
Area Î”CPD = 40 cm^{2}
(Proportional to bases BC and PC)
âˆ´ Area Î” DBC = 40 Ã— 3/2 = 60 cm^{2}.