**Excerise 7.1**

**1. In quadrilateral ACBD, AC = AD and AB bisects \(\angle\)A . Show that \(\Delta\)ABC \(\cong\) \(\Delta\)ABD. What can you say about BC and BD?**

**Sol. **

Given that,

AC = AD and AB bisects \(\angle\)A

To prove that,

\(\Delta\)ABC \(\cong\) \(\Delta\)ABD

Proof,

In \(\Delta\)ABC and \(\Delta\)ABD,

AB = AB

AC = AD

\(\angle\)CAB =\(\angle\)DAB (AB is bisector)

Therefore, \(\Delta\)ABC \(\cong\) \(\Delta\)ABD by SAS congruence condition.

BC and BD are of equal length.

**2. ABCD is a quadrilateral in which AD = BC and \(\angle\)DAB = \(\angle\)CBA . Prove that**

**(i) \(\Delta\)ABD \(\cong\) \(\Delta\)BAC
(ii) BD = AC
(iii) \(\angle\)ABD = \(\angle\)BAC.**

**Sol.**

Given that ,

AD = BC and \(\angle\) DAB = \(\angle\)CBA

(i) In \(\Delta\)ABD and \(\Delta\)BAC,

AB = BA (Common)

\(\angle\) DAB = \(\angle\)CBA (Given)

AD = BC (Given)

Therefore, \(\Delta\)ABD \(\cong\) \(\Delta\)BAC by SAS congruence condition.

(ii) Since, \(\Delta\)ABD \(\cong\) \(\Delta\)BAC

Therefore BD = AC by CPCT

(iii) Since, \(\Delta\)ABD \(\cong\) \(\Delta\)BAC

Therefore \(\angle\) ABD \(\cong\) \(\angle\)BAC by CPCT

**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 \(\Delta\)AOD and \(\Delta\)BOC,

\(\angle\)A = \(\angle\)B (Perpendicular)

\(\angle\)AOD = \(\angle\)BOC (Vertically opposite angles)

AD = BC (Given)

Therefore, \(\Delta\)AOD \(\cong\) \(\Delta\)BOC by AAS congruence condition.

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 \(\Delta\)ABC \(\cong\) \(\Delta\)CDA.**

**Sol. **

Given that,

l \(\parallel\) m and p \(\parallel\) q

To prove,

\(\Delta\)ABC ≅ \(\Delta\)CDA

Then Proof that,

In \(\Delta\)ABC and \(\Delta\)CDA,

\(\angle\)BCA =\(\angle\)DAC (Alternate interior angles)

AC = CA (Common)

\(\angle\)BAC = \(\angle\)DCA (Alternate interior angles)

Therefore, \(\Delta\)ABC \(\cong\) \(\Delta\)CDA by ASA congruence condition.

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

**(i) \(\Delta\)APB \(\cong\) \(\Delta\)AQB**

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

**Sol.**

Given,

l is the bisector of an angle \(\angle\)A.

BP and BQ are perpendiculars.

(i) In \(\Delta\)APB and \(\Delta\)AQB,

\(\angle\)P = \(\angle\)Q (Right angles)

\(\angle\)BAP =\(\angle\)BAQ (l is bisector)

AB = AB (Common)

Therefore, \(\Delta\)APB \(\cong\) \(\Delta\)AQB by AAS congruence condition.

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

**6. AC = AE, AB = AD and \(\angle\)BAD = ****\(\angle\)****EAC. Show that BC = DE.**** **

** **

**Sol.**

Given that,

AC = AE, AB = AD and \(\angle\)BAD = \(\angle\)EAC

To show,

BC = DE

then Proof that,

\(\angle\)BAD = \(\angle\)EAC (Adding \(\angle\)DAC both sides)

\(\angle\)BAD + \(\angle\)DAC = \(\angle\)EAC + \(\angle\)DAC

\(\Rightarrow\) \(\angle\)BAC = \(\angle\)EAD

In \(\Delta\)ABC and \(\Delta\)ADE,

AC = AE (Given)

\(\angle\)BAC = \(\angle\)EAD

AB = AD (Given)

Therefore, \(\Delta\)ABC \(\cong\) \(\cong\)\(\Delta\)ADE by SAS congruence condition.

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 \(\angle\)BAD = \(\angle\)ABE and \(\angle\)EPA = \(\angle\)DPB . Show that**

** (i) \(\Delta\)DAP \(\cong\) \(\Delta\)EBP**

** (ii) AD = BE**

**Sol.**

Given,

P is mid-point of AB.

\(\angle\)BAD = \(\angle\)ABE and \(\angle\)EPA = \(\angle\)DPB

(i) \(\angle\)EPA =\(\angle\)DPB (Adding \(\angle\)DPE both sides)

\(\angle\)EPA + \(\angle\)DPE = \(\angle\)DPB + \(\angle\)DPE

\(\Rightarrow\) \(\angle\)DPA = \(\angle\)EPB

In \(\Delta\)DAP \(\cong\) \(\Delta\)EBP,

\(\angle\)DPA = \(\angle\)EPB

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

\(\angle\)BAD =\(\angle\)ABE (Given)

Therefore, \(\Delta\)DAP \(\cong\) \(\Delta\)EBP by ASA congruence condition.

(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) \(\Delta\)AMC \(\cong\) \(\Delta\)BMD**

**(ii) \(\angle\)DBC is a right angle.**

**(iii) \(\Delta\)DBC \(\cong\) \(\Delta\)ACB**

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

**Sol.**

Given,

\(\angle\)C = 90°, M is the mid-point of AB and DM = CM

(i) In \(\Delta\)AMC and \(\Delta\)BMD,

AM = BM (M is the mid-point)

\(\angle\)CMA = \(\angle\)DMB (Vertically opposite angles)

CM = DM (Given)

Therefore, \(\Delta\)AMC \(\cong\) \(\Delta\)BMD by SAS congruence condition.

(ii) \(\angle\)ACM = \(\angle\)BDM (by CPCT)

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

Now,

\(\angle\)ACB + \(\angle\)DBC = 180° (co-interiors angles)

\(\Rightarrow\) 90° + \(\angle\)B = 180°

\(\Rightarrow\) \(\angle\)DBC = 90°

(iii) In \(\Delta\)DBC and \(\Delta\)ACB,

BC = CB (Common)

\(\angle\)ACB = \(\angle\)DBC (Right angles)

DB = AC

Therefore, \(\Delta\)DBC \(\cong\) \(\Delta\)ACB by SAS congruence condition.

(iv) DC = AB (\(\Delta\)DBC \(\cong\) \(\Delta\)ACB)

\(\Rightarrow\) DM = CM = AM = BM (M is mid-point)

\(\Rightarrow\) DM + CM = AM + BM

\(\Rightarrow\) CM + CM = AB

\(\Rightarrow\) CM = 1/2AB

__Exercise 7.2__

**1. In an isosceles triangle ABC, with AB = AC, the bisectors of ****\(\angle\)****B and ****\(\angle\)****C intersect each other at O. Join A to O. Show that :**

**(a) OB = OC (b) AO bisects ****\(\angle\)****A**

**Sol.**

Sol. Given that,

AB = AC, the bisectors of \(\angle\)B and \(\angle\)C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,

∴ \(\angle\)B = \(\angle\)C

⇒ 1/2\(\angle\)B = 1/2\(\angle\)C

⇒ \(\angle\)OBC = \(\angle\)OCB (Angle bisectors.)

⇒ OB = OC (Side opposite to the equal angles are equal.)

(ii) In ΔAOB and ΔAOC,

AB = AC (Given)

AO = AO (Common)

OB = OC (Proved above)

Therefore, ΔAOB ≅ ΔAOC by SSS congruence condition.

\(\angle\)BAO = \(\angle\)CAO (by CPCT)

Thus, AO bisects \(\angle\)A.

**2. In ΔABC, AD is the perpendicular bisector of BC . Show that ΔABC is an isosceles triangle in which AB = AC.**

**Sol.**

Given,

AD is the perpendicular bisector of BC

To show,

AB = AC

Proof,

In ΔADB and ΔADC,

AD = AD (Common)

∠ADB = ∠ADC

BD = CD (AD is the perpendicular bisector)

Therefore, ΔADB ≅ ΔADC by SAS congruence condition.

AB = AC

**3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.**

**Sol.**

Given that,

BE and CF are altitudes.

AC = AB

To show,

BE = CF

Proof,

In ΔAEB and ΔAFC,

∠A = ∠A (Common)

∠AEB = ∠AFC (Right angles)

AB = AC (Given)

Therefore, ΔAEB ≅ ΔAFC by AAS congruence condition.

Thus, BE = CF by CPCT.

**4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal . Show that**

** (i) ΔABE ≅ ΔACF**

** (ii) AB = AC, i.e., ABC is an isosceles triangle.**

**Sol.**

Given,

BE = CF

(i) In ΔABE and ΔACF,

∠A = ∠A (Common)

∠AEB = ∠AFC (Right angles)

BE = CF (Given)

Therefore, ΔABE ≅ ΔACF by AAS congruence condition.

(ii) Thus, AB = AC by CPCT and therefore ABC is an isosceles triangle.

**5. ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.**

**Sol.**

Sol. Given that,

ABC and DBC are two isosceles triangles.

To show,

∠ABD = ∠ACD

Proof,

In ΔABD and ΔACD,

AD = AD (Common)

AB = AC (ABC is an isosceles triangle.)

BD = CD (BCD is an isosceles triangle.)

Therefore, ΔABD ≅ ΔACD by SSS congruence condition. Thus, ∠ABD = ∠ACD by CPCT.

**6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.**

**Sol.**

Sol. Given that,

AB = AC and AD = AB

To show,

∠BCD is a right angle.

Proof,

In ΔABC,

AB = AC (Given)

⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)

In ΔACD,

AD = AB

⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)

Now,

In ΔABC,

∠CAB + ∠ACB + ∠ABC = 180°

⇒ ∠CAB + 2∠ACB = 180°

⇒ ∠CAB = 180° – 2∠ACB — (i)

Similarly in ΔADC,

∠CAD = 180° – 2∠ACD — (ii)

also,

∠CAB + ∠CAD = 180° (BD is a straight line.)

Adding (i) and (ii)

∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD

⇒ 180° = 360° – 2∠ACB – 2∠ACD

⇒ 2(∠ACB + ∠ACD) = 180°

⇒ ∠BCD = 90°

**7.ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.**

**Sol.**

Given,

∠A = 90° and AB = AC

A/q,

AB = AC

⇒ ∠B = ∠C (Angles opposite to the equal sides are equal.)

Now,

∠A + ∠B + ∠C = 180° (Sum of the interior angles of the triangle.)

⇒ 90° + 2∠B = 180°

⇒ 2∠B = 90°

⇒ ∠B = 45°

Thus, ∠B = ∠C = 45°

**8. Show that the angles of an equilateral triangle are 60° each.**

**Sol.**

Let ABC be an equilateral triangle.

BC = AC = AB (Length of all sides is same)

⇒ ∠A = ∠B = ∠C (Sides opposite to the equal angles are equal.)

Also,

∠A + ∠B + ∠C = 180°

⇒ 3∠A = 180°

⇒ ∠A = 60°

Therefore, ∠A = ∠B = ∠C = 60°

Thus, the angles of an equilateral triangle are 60° each.