**Exercise: 7.1 (Page No: 118)**

**1. In quadrilateral ACBD, AC = AD and AB bisect A (see Fig. 7.16). Show that ΔABC ΔABD. What can you say about BC and BD?**

**Solution:**

It is given that AC and AD are equal i.e. AC = AD and the line segment AB bisects A.

We will have to now prove that the two triangles ABC and ABD are similar i.e. **ΔABC ΔABD**

**Proof:**

Consider the triangles ΔABC and ΔABD,

(i) AC = AD (It is given in the question)

(ii) AB = AB (Common)

(iii) CAB = DAB (Since AB is the bisector of angle A)

So, by **SAS congruency criterion**, ΔABC ΔABD.

For the 2^{nd} part of the question, BC and BD are of equal lengths by the rule of C.P.C.T.

**2. ABCD is a quadrilateral in which AD = BC and DAB = CBA (see Fig. 7.17). Prove that**

(i) ΔABD ΔBAC

(ii) BD = AC

(iii) ABD = BAC.

**Solution:**

The given parameters from the questions are DAB = CBA and AD = BC.

(i) ΔABD and ΔBAC are similar by SAS congruency as

AB = BA (It is the common arm)

DAB = CBA and AD = BC (These are given in the question)

So, triangles ABD and BAC are similar i.e. ΔABD ΔBAC. (Hence proved).

(ii) It is now known that ΔABD ΔBAC so,

BD = AC (by the rule of CPCT).

(iii) Since ΔABD ΔBAC so,

Angles ABD = BAC (by the rule of CPCT).

**3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.**

**Solution:**

It is given that AD and BC are two equal perpendiculars to AB.

We will have to prove that **CD is the bisector of AB**

Now,

Triangles ΔAOD and ΔBOC are similar by AAS congruency since:

(i) A = B (They are perpendiculars)

(ii) AD = BC (As given in the question)

(iii) AOD = BOC (They are vertically opposite angles)

∴ ΔAOD ΔBOC.

So, AO = OB (by the rule of CPCT).

Thus, CD bisects AB (Hence proved).

**4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ΔCDA.**

**Solution:**

It is given that p q and l m

**To prove:**

Triangles ABC and CDA are similar i.e. ΔABC ΔCDA

**Proof**:

Consider the ΔABC and ΔCDA,

(i) BCA = DAC and BAC = DCA Since they are alternate interior angles

(ii) AC = CA as it is the common arm

So, by **ASA congruency criterion, **ΔABC ΔCDA.

**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 (see Fig. 7.20). Show that:**

(i) ΔAPB ΔAQB (ii) BP = BQ or B is equidistant from the arms of A.

**Solution:**

It is given that the line “*l*” is the bisector of angle A and the line segments BP and BQ are perpendiculars drawn from *l*.

(i) ΔAPB and ΔAQB are similar by AAS congruency because:

P = Q (They are the two right angles)

AB = AB (It is the common arm)

BAP = BAQ (As line *l *is the bisector of angle A)

So, ΔAPB ΔAQB.

(ii) By the rule of CPCT, BP = BQ. So, it can be said the point B is equidistant from the arms of A.

**6. In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE.**

**Solution:**

It is given in the question that AB = AD, AC = AE, and ∠BAD = ∠EAC

**To prove:**

The line segment BC and DE are similar i.e. BC = DE

**Proof:**

We know that BAD = EAC

Now, by adding DAC on both sides we get,

BAD + DAC = EAC +DAC

This implies, BAC = EAD

Now, ΔABC and ΔADE are similar by SAS congruency since:

(i) AC = AE (As given in the question)

(ii) BAC = EAD

(iii) AB = AD (It is also given in the question)

∴ Triangles ABC and ADE are similar i.e. ΔABC ΔADE.

So, by the rule of CPCT, it can be said that BC = DE.

**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. 7.22). Show that**

**(i) ΔDAP ΔEBP**

**(ii) AD = BE**

**Solutions:**

In the question, it is given that P is the mid-point of line segment AB. Also, BAD = ABE and EPA = DPB

(i) It is given that EPA = DPB

Now, add DPE on both sides,

EPA +DPE = DPB+DPE

This implies that angles DPA and EPB are equal i.e. DPA = EPB

Now, consider the triangles DAP and EBP.

DPA = EPB

AP = BP (Since P is the mid-point of the line segment AB)

BAD = ABE (As given in the question)

So, by **ASA congruency**, ΔDAP ΔEBP.

(ii) By the rule of CPCT, AD = BE.

**8. 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 Fig. 7.23). Show that:**

(i) ΔAMC ΔBMD

(ii) DBC is a right angle.

(iii) ΔDBC ΔACB

(iv) CM = ½ AB

**Solution:**

It is given that M is the mid-point of the line segment AB, C = 90°, and DM = CM

(i) Consider the triangles ΔAMC and ΔBMD:

AM = BM (Since M is the mid-point)

CM = DM (Given in the question)

CMA = DMB (They are vertically opposite angles)

So, by **SAS congruency criterion**, ΔAMC ΔBMD.

(ii) ACM = BDM (by CPCT)

∴ AC BD as alternate interior angles are equal.

Now, ACB +DBC = 180° (Since they are co-interiors angles)

⇒ 90° +B = 180°

∴ DBC = 90°

(iii) In ΔDBC and ΔACB,

BC = CB (Common side)

ACB = DBC (They are right angles)

DB = AC (by CPCT)

So, ΔDBC ΔACB by **SAS congruency**.

(iv) DC = AB (Since ΔDBC ΔACB)

⇒ DM = CM = AM = BM (Since M the is mid-point)

So, DM + CM = BM+AM

Hence, CM + CM = AB

⇒ CM = (½) AB