**1. In quadrilateral ACBD, AC = AD and AB bisects ∠A . Show that ΔABC ≅ ΔABD. What can you say about BC and BD?**

**Sol. **

Given that,

AC = AD and AB bisects

To prove that,

Proof,

In

AB = AB

AC = AD

Therefore,

BC and BD are of equal length.

**2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA . Prove that**

**(i) ΔABD ≅ ΔBAC**

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

**Sol.**

Given that ,

AD = BC and

(i) In

AB = BA (Common)

AD = BC (Given)

Therefore,

(ii) Since,

Therefore BD = AC by CPCT

(iii) Since,

Therefore

**3. AD and BC are equal perpendiculars to a line segment AB . Show that CD bisects AB.**** **

** **

**Sol. **

Given that,

AD and BC are equal perpendiculars to AB.

To prove,

CD bisects AB

Proof,

In

AD = BC (Given)

Therefore,

Now,

AO = OB (CPCT). CD bisects AB.

**4. l and m are two parallel lines intersected by another pair of parallel lines p and q . Show that ΔABC ≅ ΔCDA.**

**Sol. **

Given that,

l

To prove,

Then Proof that,

In

AC = CA (Common)

Therefore,

**5. 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 ∠A . Show that:**

**(i) ΔAPB ≅ ΔAQB**

**(ii) BP = BQ or B is equidistant from the arms of ∠A.**

**Sol.**

Given,

l is the bisector of an angle

BP and BQ are perpendiculars.

(i) In

AB = AB (Common)

Therefore,

(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of

**6. AC = AE, AB = AD and ∠BAD = **

∠

**EAC. Show that BC = DE.**

** **

**Sol.**

Given that,

AC = AE, AB = AD and

To show,

BC = DE

then Proof that,

In

AC = AE (Given)

AB = AD (Given)

Therefore,

BC = DE by CPCT.

**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 . Show that**

**(i)**Δ DAP ≅ Δ EBP

**(ii) AD = BE**

**Sol.**

Given,

P is mid-point of AB.

(i)

In

AP = BP (P is mid-point of AB)

Therefore,

(ii) AD = BE by CPCT.

**8. A 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. Show that:**

**(i) ΔAMC ≅ ΔBMD**

**(ii) ∠DBC is a right angle.**

**(iii) ΔDBC ≅ ΔACB**

**(iv) CM = 1/2 AB**

**Sol.**

Given,

(i) In

AM = BM (M is the mid-point)

CM = DM (Given)

Therefore,

(ii)

Therefore, AC || BD as alternate interior angles are equal.

Now,

(iii) In

BC = CB (Common)

DB = AC

Therefore,

(iv) DC = AB (