# Question-1

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

**Solution:**

In a parallelogram ABCD, âˆ 1 = âˆ 2, and âˆ 3 = âˆ 4

**To Prove:**âˆ APB = 90Â°

**Proof:**Since AD Ã¯Ã¯ BC with transversal AB,

.

^{.}. âˆ A + âˆ B = 180Â°(Consecutive Interior angles) â‡’âˆ A + âˆ B = 90Â° â‡’ âˆ 2 + âˆ 3 = 90Â° ...... (i)

Now in Î”ABP,

we have

âˆ 2 + âˆ 3 + âˆ APB = 180Â° ...... (ii) â‡’ 90Â° + âˆ APB = 180Â°

.

^{.}. âˆ APB = 180Â° - 90Â° = 90Â°

# 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 âˆ 1 = âˆ 2 and diagonals AC and DB intersect at P.

**To Prove:**âˆ 3 = âˆ 4 and âˆ APB = 90Â°

**Proof**: Since AD Ã¯Ã¯ BC and DB is the transversal

.

^{.}. âˆ 2 = âˆ 4 (alternate interior angles )......(i)

Similarly,

âˆ 1 = âˆ 3 ..... (ii)

But âˆ 1 = âˆ 2 (given) ...(iii)

From equations (i), (ii) and (iii), we get,

âˆ 3 = âˆ 4

Thus diagonal DB bisects âˆ B also.

Now in Î”ABP and Î”ADP, âˆ 2 = âˆ 3

AB = AD..... (sides opposite to equal angles)

DP = PB .... (diagonals bisects each other)

and AP = AP ... (common)

Î”ABP â‰… Î”ADP

.

^{.}. âˆ APD = âˆ APB ...... (c.p.c.t.)

But âˆ APD + âˆ APB = 180Â° ......(linear pair)

2 âˆ APB = 180Â° â‡’âˆ APB = 90Â°

**âˆ´**Diagonals are bisect each other at 90Â°.

# Question-3

**In a****Î”ABC, 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 Ã¯Ã¯ RQ, AB Ã¯Ã¯ QP and AC Ã¯Ã¯ RP

**To Prove:**BC = QR

**Proof:**Since QC Ã¯Ã¯ AB and QA Ã¯Ã¯ 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 Ã¯Ã¯ DC and BC and DE are transversals,

.

^{.}. âˆ 1 = âˆ 2 and âˆ 3 = âˆ 4

(Alternate Interior angles of parallel sides)

Now in Î”BEP and Î”CDP,

we have

âˆ 2 = âˆ 1

âˆ 3 = âˆ 4

and BE = DC

Î”BEP â‰… Î”CDP ...... (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) âˆ’ (2) â‡’ AB - AL = DC - CM â‡’ LB = DM

Now in Î”DMP and Î”BLP,

we have

âˆ 1 = âˆ 4...... (Alternate Interior angles)

âˆ 2 = âˆ 3...... (Alternate Interior angles)

and DM = BL

.

^{.}. Î”DMP â‰… Î”BLP

.

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

.

^{.}. BD and ML bisect each other.

# Question-6

**Î”ABC and Î”DEF 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 Î”ABC and Î”DEF,

AB = DE and AB Ã¯Ã¯ DE

BC = EF and BC Ã¯Ã¯ EF

**To Prove:**AC = DF and AC Ã¯Ã¯ DF

**Construction:**Join AD, CF and BE

**Proof**: Since AB = DE and AB Ã¯Ã¯ DE

.

^{.}. Quadrilateral ABED is a parallelogram (one pair of opposite sides are equal and parallel)

.

^{.}. AD = BE and AD Ã¯Ã¯ BE .....(i)

Similarly quadrilateral BEFC is a parallelogram

.

^{.}. BE = CF and BE Ã¯Ã¯ CF......(ii)

From (i) and (ii) we get,

AD = CF and AD Ã¯Ã¯ CF

.

^{.}. ADFC is a parallelogram

.

^{.}. AC = DF and AC Ã¯Ã¯ 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 Ã¯Ã¯ CD with GH as transversal

.

^{.}. âˆ BGH + âˆ DHG = 180Â°

.

^{.}. âˆ BGH + âˆ DHG = x 180Â° â‡’ âˆ 2 + âˆ 3 = 90Â°..... (i)

Now, âˆ 2 + âˆ 3 + âˆ M = 180Â°.....(ii) â‡’ 90Â° + âˆ M = 180Â°

.

^{.}. âˆ M = 90Â°

Similarly âˆ N = 90Â°

Now âˆ HGB + âˆ HGA = 180Â° ......(linear pair)

.

^{.}. âˆ HGB + âˆ HGA = x 180Â°

.

^{.}. âˆ 1 + âˆ 2 = 90Â°......(iii)

.

^{.}. âˆ NGM = 90Â°

Similarly âˆ NHM = 90Â°

Since all the four angles of quadrilateral NGMH is 90Â°

.

^{.}. 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 Ã¯Ã¯ DC...... (opp. sides of a Ã¯Ã¯

^{gm})

XB Ã¯Ã¯ DY.....(ii)

Since in quadrilateral XBYD, XB = DY and XB Ã¯Ã¯ DY

.

^{.}. XBYD is a parallelogram.

.

^{.}. DX Ã¯Ã¯ YB â‡’ PX Ã¯Ã¯ YQ......(iii)

Similarly we can prove,PY Ã¯Ã¯ 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 Î”ABC, E is the mid-point of AB and F is the mid-point of BC.

.

^{.}. EF = AC and EF Ã¯Ã¯ AC...(i)

Similarly, HG = AC and HG Ã¯Ã¯ AC...(ii)

From eqns (i) and (ii), we get

HG Ã¯Ã¯ 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 Î”ABC, D, E and F are the mid-points of sides AB, BC and AC respectively.

**To Prove:**Î”ADF â‰… Î”DBE â‰… Î”ECF â‰… Î”FDE

**Proof:**In Î”ABC, F is the mid-point of AC and D is

the mid-point of AB.

.

^{.}. FD = CB and FD Ã¯Ã¯ CB (mid-point theorem) â‡’ FD = CE and FD Ã¯Ã¯ CE...(i)

Similarly,

DE = FC and DE Ã¯Ã¯ FC..(ii)

and FE = DB and FE Ã¯Ã¯ 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.

.

^{.}. Î”DEF â‰… Î”ADF..(iv)

.

^{.}. Î”DEF â‰… Î”DBE...(v)

.

^{.}. Î”DEF â‰… Î”FEC ...(vi)

From (iv), (v) and (vi) we get,

Î”ADF â‰… Î”DBE â‰… Î”ECF â‰… Î”DEF

# Question-11

**In****Î”ABC, âˆ B = 90Â° and P is the mid-point of AC. Prove that PB = PA =****AC.**

**Solution:**

In the Î”ABC, P is the mid-point of AC and âˆ B = 90Â°.

**To Prove:**PA = PB = AC

**Construction**: Draw PK Ã¯Ã¯ BC

**Proof:**Since KP Ã¯Ã¯ BC with transversal AB

.

^{.}. âˆ 1 = âˆ B = 90Â°

Also, âˆ 1 + âˆ 2 = 180Â°

.

^{.}. âˆ 2 = 180Â° - âˆ 1 = 180Â° - 90Â° = 90Â°

Now in Î”APK and Î”BPK,

KP = KP...(common)

âˆ 1 = âˆ 2...(prove above)

AK = KB...(since KP Ã¯Ã¯ BC and P is the mid-point of AC)

.

^{.}. Î”APK â‰… Î”BPK

.

^{.}. PA = PB but

PA = AC ...( P is the mid-point of AC)

PA = PB = AC.

# Question-12

**In****Î”ABC, 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 Î”ABC, D, E, F are mid-points of BC, AB and AC.

Also, AB = AC

**To Prove:**AM = MD and Î”AMF = 90Â°

**Construction:**Join ED and FD.

**Proof:**In Î”ABC, AB = AC â‡’ AB = AC â‡’ AE = AF...(i)

Now in the Î”ABC, E is the mid-point of AB and F is the mid-point of AC.

.

^{.}. EF = BC and EF Ã¯Ã¯ BC

In Î”ABC, E is the mid-point of AB and D is the mid-point of BC.

.

^{.}. ED = AC and ED Ã¯Ã¯ AC â‡’ ED = AF and ED Ã¯Ã¯ 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 âŠ¥ EF

# Question-13

**In the****Î”ABC, 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:**Î”ABC in which E is the mid-point of the median AD.

**To Prove:**AF = AC

**Construction:**Draw DG Ã¯Ã¯ BF, cutting AC at G.

**Proof:**In the Î”ADG, EF Ã¯Ã¯ 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 Î”BCF, DG Ã¯Ã¯ 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 Î”ABC, P and Q are the mid-points of AB and BC respectively.

.

^{.}. PQ Ã¯Ã¯ AC and PQ = AC........(i)

In the Î”ADC, R and S are the mid-points of CD and AD respectively.

.

^{.}. RS Ã¯Ã¯ AC and RS = AC........(ii)

From (i) and (ii), we have

PQ Ã¯Ã¯ 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 Î”ABD, P is the mid-points of AB and S is the mid-point of AD.

.

^{.}. PS Ã¯Ã¯ BD â‡’ PN Ã¯Ã¯ MO

Also, from (i), PQ Ã¯Ã¯ AC â‡’ PM Ã¯Ã¯ NO

Thus in quadrilateral PMON,

we have

PN Ã¯Ã¯ MO and PM Ã¯Ã¯ NO â‡’ PMON is a parallelogram â‡’ âˆ MPN = âˆ MON [opp. angles of parallelogram

^{ }are equal]

â‡’ âˆ MPN = âˆ BOA [

^{.}.

^{.}âˆ BOA = âˆ MON] â‡’ âˆ MPN = 90Â° [

^{.}.

^{.}AC âŠ¥ BD,

^{.}.

^{.}âˆ BOA = 90Â°]

â‡’ âˆ QPS = 90Â° [

^{.}.

^{.}âˆ MPN = âˆ QPS]

Thus, PQRS is a parallelogram

whose one angle âˆ QPS = 90Â°.

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 Î”ADG, E is the mid-point of AD and EC Ã¯Ã¯ AG

.

^{.}. EC = AG and EC Ã¯Ã¯ AG â‡’ 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 Ã¯Ã¯ FC and AF Ã¯Ã¯ EC,

.

^{.}. Quadrilateral AFCE is a parallelogram

.

^{.}. EC = AF...(iii)

From equations (i) and (iii) we get,

2AF = AG â‡’ AG = 2AF