**Question 1:**

*In quadrilateral PQRS, PR = PQ and PQ bisects ∠P (look at the fig.). Show that the △PQR = △PQS. What can you say about QR and QS?*

**Solution:**

In

PR = PS

PQ = PQ (common)

Hence Proved.

Therefore, QR = QS (CPCT)

*Question 2:*

*ABCD is a quadrilateral in which AD = BC and ∠DAB= ∠CBA (see Fig.). Prove that*

*(i) △ABD=△BAC*

*(ii) BD = AC*

*(iii) ∠ABD=∠BAC*

**Solution:**

In the given figure, ABCD is a quadrilateral in which AD = BC and

In

AD = BC [Given]

AB = AB [Common]

BD = AC [CPCT]

and

Hence Proved.

**Question 3:**

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

**Solution:**

In

AD = BC [Given]

Also, AO = BO [CPCT]

Hence, CD bisects AB Proved.

**Question 4:**

*L and m are two para;;e; lines intersected by another pair of parallel lines p and q (see fig.). Show that △ABC+△CDA .*

**Solution:**

In the given figure, ABCD is a parallelogram in which AC is a diagonal, i.e., AB || DC and BC || AD.

In

AC = AC (Common)

Hence Proved.

**Question 5:**

*Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendicular from B to the arms of ∠A (See fig.). Show that:*

(i)

(ii) BP = BQ or B is equidistant from the arms of

**Solution:**

In

AB = AB

Also, BP = BQ (By CPCT)

i.e, B is equidistant from the arms of

Hence proved.

**Question 6:**

*In the fig, AC = AE, AB = AD and △BAD=△EAC . Show that BC = DE.*

**Solution:**

Now, in

AB = AD

AC = AE

BC = DE (CPCT)

Hence Proved.

**Question 7:**

*AB is a line segment and p is its mid-point. D and E are points on the same side of AB such that ∠BAD=∠ABE and ∠EPA=∠DPB (see fig.). Show that*

* (i) △DAP=△EBP*

* (ii) AD = BE*

*Solution:*

In

AP = BP (P is the midpoint of the line segment AB)

AD = BE (ASA)

AD = BE

Hence Proved.

**Question 8:**

*In right triangle ABC, right angles 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 fig.). Show that :*

* (i) △AMC=△BMD*

* (II) ∠DBC is a right angle.*

* (iii) △DBC=△ACB*

* (iv) CM=12AB*

**Solution:**

In

(i) DM = CM (given)

BM = AM (M is the midpoint of AB)

Hence Proved.

(ii) AC || BD

Hence proved.

(iii) In

DB = AC (CPCT)

BC = BC (Common)

Hence proved.

(iv) AB = CD

Hence,