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Question-1

In a parallelogram ABCD, bisectors of consecutive angles A and B intersect at P. Prove that APB = 90°.

 

Solution:
In a parallelogram ABCD, 1 = 2, and 3 = 4

To Prove:
APB = 90°

Proof: Since AD
ïï BC with transversal AB, 

..
A + B = 180°(Consecutive Interior angles) A + B = 90° 2 + 3 = 90°  ...... (i)
Now in
ΔABP,
we have
 
2 + 3 + APB = 180° ...... (ii)  90° + APB = 180°

..
APB = 180° - 90° = 90°

Question-2

If a diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle and then the two diagonals are perpendicular to each other.
 

Solution:
Given: A parallelogram ABCD, in which 1 = 2 and diagonals AC and DB intersect at P. 

To Prove: 3 = 4 and APB = 90°

Proof: Since AD ïï BC and DB is the transversal
..
2 = 4  (alternate interior angles )......(i) 

Similarly,
1 = 3            ..... (ii)
But 1 = 2 (given)  ...(iii)
From equations (i), (ii) and (iii), we get,
 3 = 4
Thus diagonal DB bisects B also. 
Now in ΔABP and ΔADP,  2 = 3
AB = AD..... (sides opposite to equal angles)
DP = PB .... (diagonals bisects each other)
and AP = AP ... (common)

ΔABP ΔADP
...
APD = APB          ...... (c.p.c.t.)
But
APD + APB = 180° ......(linear pair)
2
APB = 180° APB = 90°
 Diagonals are bisect each other at 90°.

Question-3

In a ΔABC, lines are drawn through A, B and C parallel to sides BC, CA and AB respectively forming a triangle PQR. Prove that BC = QR.

Solution:
BC ïï RQ, AB ïï QP and AC ïï RP

To Prove: BC = QR

Proof: Since QC
ïï AB and QA ïï BC

... ABCQ is a parallelogram (opposite sides are parallel)

... BC = AQ     ...... (opposite sides of parallelogram
)....(i)

Similarly BCAR is a parallelogram and BC = AR ............(ii)
By adding (i) and (ii) 
we get
2BC = AQ + AR = QR
... BC = QR. 

Question-4

ABCD is a parallelogram AB is produced to E so that BE = AB. Show that ED bisects BC.


 

Solution:
AB = BE    ..... (given)
Also AB = DC   ......(opposite sides of a parallelogram
)
... BE = DC (Since AB = DC and AB = BE )

Since AE
ïï DC and BC and DE are transversals,

..
1 = 2 and 3 = 4
(Alternate Interior angles of parallel
 sides)
Now in
ΔBEP and ΔCDP, 
we have
        
2 = 1
        
3 = 4
and    BE = DC
 
ΔBEP ΔCDP    ...... (ASA Criterion)

... CP = PB        (c.p.c.t.)

... ED bisects BC at P.

Question-5

ABCD is a parallelogram L and M are points on AB and DC respectively such that AL = CM. Show that BD and ML bisect each other.

Solution:
 


Given: In
a parallelogram ABCD, AL = CM
To Prove: DP = PB and MP = PL
Proof: Since DC and AB are two opposite sides of parallelogram
 ABCD
... AB = DC  .............(1)
Also AL = CM  ...........(2) (1) − (2) ⇒ AB - AL = DC - CM LB = DM
Now in
ΔDMP and ΔBLP, 
we have

1 = 4...... (Alternate Interior angles)
2 = 3...... (Alternate Interior angles)
and DM = BL

..
ΔDMP ΔBLP

... DP = PB and MP = PL....(c.p.c.t.)


... BD and ML bisect each other.

Question-6

ΔABC and ΔDEF are two triangles such that AB, BC are respectively equal and parallel to DE, EF; show that AC is equal and parallel to DF.


Solution:
In the ΔABC and ΔDEF,
AB = DE and AB
ïï DE
BC = EF and BC
ïï EF

To Prove:   AC = DF and AC
ïï DF

Construction: Join AD, CF and BE

Proof: Since AB = DE and AB
ïï DE
... Quadrilateral ABED is a parallelogram (one pair of opposite sides are equal and parallel)
... AD = BE and AD
ïï BE .....(i)
Similarly quadrilateral BEFC is a parallelogram
... BE = CF and BE
ïï CF......(ii)
From (i) and (ii) we get,  
AD = CF and AD
ïï CF
... ADFC is a parallelogram
... AC = DF and AC
ïï DF

Question-7

AB, CD are two parallel lines and a transversal l intersects AB at G and CD at H. Prove that the bisectors of the interior angles form a parallelogram, with all the angles are right angles.


Solution:

Given: AB and CD are two parallel lines and the transversal l cut AB at G and CD at H. The bisectors of interior angles intersect at M and N.

To Prove: Quadrilateral NGMH is a rectangle

Proof: AB
ïï CD with GH as transversal
...
BGH + DHG = 180°
...
BGH + DHG = x 180° ⇒ ∠2 + 3 = 90°..... (i)
Now, 
2 + 3 + M = 180°.....(ii) 90° + M = 180°
..
M = 90°

Similarly
N = 90°

Now
HGB + HGA = 180°  ......(linear pair)
...
HGB + HGA = x 180°
..
1 + 2 = 90°......(iii)
...
NGM = 90°

Similarly
NHM = 90°

Since all the four angles of quadrilateral NGMH is 90°
... Quadrilateral NGMH is a parallelogram and rectangle.

Question-8

X and Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined to intersect in P. CX and BY are joined to intersect in Q. Show that PXQY is a parallelogram.


Solution:
Given: X and Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD.
AX = XB and DY = YC

To Prove: PXQY is a parallelogram.

Proof: AB = DC
..AB =   DC
XB = DY......(i)

Also AB
ïï DC...... (opp. sides of a ïïgm)
       XB
ïï DY.....(ii)

Since in quadrilateral XBYD, XB = DY and XB
ïï DY
... XBYD is a parallelogram.
... DX
ïï YB  PX ïï YQ......(iii)

Similarly we can prove,PY
ïï XQ ......(iv)
From (iii) and (iv), we get PXQY is a parallelogram.

Question-9

Prove that the quadrilateral formed by joining the mid-points of the pairs of consecutive sides of a quadrilateral is a parallelogram.


 

Solution:

Given: In quadrilateral ABCD, E, F, G and H are the mid-points of AB, BC, CD and DA respectively.

To Prove: EFGH is a parallelogram

Construction: Join AC

Proof: In
ΔABC, E is the mid-point of AB and F is the mid-point of BC.
... EF = AC and EF
ïï AC...(i)

Similarly, HG = AC and HG
ïï AC...(ii)
From eqns (i) and (ii), we get
HG
ïï EF and HG = EF ...(iii)
... EFGH is a parallelogram (One pair of opposite sides of a quadrilateral are equal and parallel.)

Question-10

Prove that four triangles formed by joining the mid-points of the sides of a triangle are congruent to each other.


 

Solution:
In ΔABC, D, E and F are the mid-points of sides AB, BC and AC respectively.

To Prove:
ΔADF ΔDBE ΔECF ΔFDE

Proof: In
ΔABC, F is the mid-point of AC and D is
the mid-point of AB.
... FD = CB and FD
ïï CB (mid-point theorem)  FD = CE and FD ïï CE...(i)

Similarly,
DE = FC and DE
ïï FC..(ii)
and FE = DB and FE
ïï DB...(iii)
... From (i), (ii) and (iii), we get

ADEF, DBEF and DECF are parallelograms. 
The diagonal of a parallelogram divide it into two triangles which will be congruent to each other.
...
ΔDEF ΔADF..(iv)
...
ΔDEF ΔDBE...(v)
...
ΔDEF ΔFEC ...(vi)
From (iv), (v) and (vi) we get, 

ΔADF ΔDBE ΔECF ΔDEF

Question-11

In ΔABC, B = 90° and P is the mid-point of AC. Prove that PB = PA =   AC.

Solution:
In the ΔABC, P is the mid-point of AC and B = 90°. 

To Prove: PA = PB = AC

Construction: Draw PK
ïï BC

Proof: Since KP
ïï BC with transversal AB
..
1 = B = 90°
Also,
1 + 2 = 180°
..
2 = 180° - 1 = 180° - 90° = 90°

Now in
ΔAPK and ΔBPK,
KP = KP...(common)

1 = 2...(prove above)
AK = KB...(since KP
ïï BC and P is the mid-point of AC)
...
ΔAPK ΔBPK
... PA = PB but
PA = AC ...( P is the mid-point of AC)
PA = PB = AC.

Question-12

In ΔABC, AB = AC. D, E and F are respectively the mid-points of sides BC, AB and AC. Show that line segment AD is perpendicular to the line segment EF and is bisected by it.

 

Solution:
Given: In ΔABC, D, E, F are mid-points of BC, AB and AC.
          Also, AB = AC

To Prove:  AM = MD and
ΔAMF = 90°

Construction: Join ED and FD.

Proof: In
ΔABC, AB = AC AB = AC AE = AF...(i)
Now in the
ΔABC, E is the mid-point of AB and F is the mid-point of AC.
... EF = BC and EF
ïï BC

In
ΔABC, E is the mid-point of AB and D is the mid-point of BC.
... ED = AC and ED
ïï AC  ED = AF and ED ïï AF
Similarly, DF = AE and DF || AE
... AFDE is a parallelogram

In a parallelogram, opposite sides are equal
Also,      
AE = AF ... ( AB = AC)
... AFDE is a rhombus.

The diagonals of a rhombus bisect each other perpendicularly.
... AM = MD and EM = MF
and AD
EF

Question-13

In the ΔABC, AD is the median through A and E is the mid-point of AD. BE produced meets AC at F. Prove that AF = AC.

 

Solution:
Given: ΔABC in which E is the mid-point of the median AD. 

To Prove: AF =
AC

Construction: Draw DG ïï BF, cutting AC at G.

Proof: In the
ΔADG, EF ïï DG and E is the
mid-point of AD.

... F is the mid-point of AG (mid-point theorem)
... AF = FG...(i)

Now in the
ΔBCF, DG ïï BF and D is the mid-point of BC.
... G is the mid-point of CF (mid-point theorem)
... FG = GC...(ii)

From Eqns (i) and (ii),
we get
AF = FG = GC

Since AC = AF + FG + GC
... AC = 3AF
... AF = AC

Question-14

The diagonals of a quadrilateral ABCD are perpendicular. Show that the quadrilateral, formed by joining the mid-points of its sides, is a rectangle.

 
 

Solution:

Given: A quadrilateral whose diagonals AC and BD are perpendicular to each other. P, Q, R, S are the mid-points of sides. AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.

To Prove: PQRS is a rectangle.

Proof: In
ΔABC, P and Q are the mid-points of AB and BC respectively.
... PQ
ïï AC and PQ = AC........(i)
In the
ΔADC, R and S are the mid-points of CD and AD respectively.
... RS
ïï AC and RS = AC........(ii)

From (i) and (ii), we have
PQ
ïï RS and PQ = RS
Thus, in quadrilateral PQRS, a pair of opposite sides are equal and parallel. So, PQRS is a parallelogram. Suppose the diagonals AC and BD of quadrilateral ABCD intersect at O. Now in
ΔABD, P is the mid-points of AB and S is the mid-point of AD.

... PS
ïï BD  PN ïï MO
Also, from (i), PQ
ïï AC PM ïï NO
Thus in quadrilateral PMON, 
we have

PN
ïï MO and PM ïï NO  PMON is a parallelogram ⇒ ∠MPN = MON [opp. angles of parallelogram are equal]
⇒ ∠MPN = BOA [... BOA = MON]  MPN = 90°     [...AC BD, ... BOA = 90°]
QPS = 90°     [...MPN = QPS]

Thus, PQRS is a parallelogram
whose one angle
QPS = 90°.
Hence, PQRS is a rectangle.

Question-15

E is the mid-point of side AD of a parallelogram ABCD. A line through A parallel to EC meets BC at F and DC produced at G. Prove that
(i) DG = 2AB
(ii) AG = 2AF

Solution:



In
ΔADG, E is the mid-point of AD and EC ïï AG
... EC = AG and EC
ïï AG   2EC = AG ...(i)
and C is the mid-point of DG.
... DG = DC + CG = 2DC...(ii)
but, DC = AB ...(opposite sides of a parallelogram
)
... DG = 2AB  

(ii) In AFCE, AE ïï FC and AF ïï EC,
... Quadrilateral AFCE is a
parallelogram 
... EC = AF...(iii)
From equations (i) and (iii) we get,
2AF = AG
 AG = 2AF




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