**Q.1Â Â Â In quadrilateral Â ACBD, AC = AD and AB bisects (see figure). Show that .Â ****What can you say about BC and BD? **

Now in ABC and ABD, we have Â

AC = AD [Given]

[Since AB bisects ]

and AB = AB [Common]

Therefore by SAS congruence criterion, we have

BC = BD [Since corresponding parts of congruent triangles are equal]

**Q.2Â Â Â Â Â ABCD is a quadrilateral in which AD = BC and (see figure). Prove thatÂ Â Â Â Â Â Â Â Â Â Â Â **

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In ABD and BAC, we have

AD = BC [Given]

[Given]

AB = AB [Common]

Therefore by SAS criterion of congruence, we have

, which proves (i)

BD = AC

and , which proves (ii) and (iii)

[Since corresponding parts of congruent triangles are equal]

**Q.3Â Â Â Â Â AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB. **

Since AB and CD intersect at O. Therefore ,

.... (1) [Vertically opp. angles]

In AOD and BOC, we have

[From (1)]

[Each = 90Âº]

and , AD = BC [Given]

Therefore by AAS congruence criterion, we have

OA = OB

[Since corresponding parts of congruent triangles are equal]

i.e., O is the mid- point of Â AB.

Hence, CD bisects AB.

**Q.4Â Â Â Â Â l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that .**

Since l and m are two parallel lines intersected by another pair of parallel lines p and q. Therefore , AD || BC and AB || CD

ABCD is a Â parallelogram.

i.e. ,Â AB = CD

and BC = AD

Now, in ABC and CDA, we have

AB = CD [Proved above]

BC = AD [Proved above]

and AC = AC [Common]

Therefore By SSS criterion of congruence.

**Q.5Â Â Â Â Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of (see figure). Show that :Â Â Â Â Â Â Â Â Â Â Â **

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In APB and AQB we have

[Since Each = 90Âº]

[Since AB bisects ]

AB = AB [Common]

By AAS congruence criterion, we have

, which proves (i)

BP = PQ

[Since Corresponding parts of congruent triangles are equal]

i.e., B is equidistant from the arms of which proves (ii).

**Q.6Â Â Â Â Â Â In figure AC = AE, AB = AD and Show that BC = DE. **

In ABC and ADE, we have

AB = AD [Given]

[Since

]

and , AC = AEÂ Â [Given ]

Therefore By SAS criterion of congruence, we have

BC = DE

[Since Corresponding parts of congruent triangles are equal]

**Q.7Â Â Â Â Â AB is a line segment and P is its mid- point. D and E are points on the same side of AB such that andÂ **** (see figure). Show that -Â Â Â Â Â Â Â Â Â Â Â Â (i) Â Â Â Â Â Â Â Â Â Â Â (ii) AD = BE.
**

We have ,

Now , in EBP and DAP, we have

[From (1)]

BP = AP Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Given]

and , [Given]

So, by ASA criterion of congruence, we have

BE = AD i.e., AD = BE

[Since Corresponding parts of congruent triangles are equal. ]

**Q.8Â Â Â Â In right triangle ABC, right angled at C, M is the mid- point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D ****is joined to point B (see figure). Show that : Â Â Â Â Â Â Â Â Â Â Â **

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**(i)** In AMC and BMD, we have

AM = BM [Since M is the mid- point of AB]

[Vertically opp. ]

and CM = MD [Given]

Therefore By SAS criterion of congruence , we have

**(ii)** Now,

... (1)

[Since corresponding parts of congruent triangles are equal]

Thus, transversal CD cuts CA and BD at C and D respectively such that the alternate angles are equal. Therefore , BD || CA.

= 180Âº

[Since sum of consecutive interior angles are supplementary ]

90Âº = 180Âº [Since = 90Âº]

= 180Âº - 90Âº

90Âº

**(iii)** Now, in DBC and ACB, we have

BD = CA [From (1)]

[Since Each = 90Âº]

BC = BC [Common]

Therefore by SAS criterion of congruence, we have

**(iv)** CD = AB [Since corresponding parts of congruent triangles are equal]

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