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**Exercise 7.1****1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?**

**Answer**

Given,

AC = AD and AB bisects ∠A

To prove,

ΔABC ≅ ΔABD

Proof,

In ΔABC and ΔABD,

AB = AB (Common)

AC = AD (Given)

∠CAB = ∠DAB (AB is bisector)

Therefore, ΔABC ≅ ΔABD by SAS congruence condition.

BC and BD are of equal length.

**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.**

**Answer**

Given,

AD = BC and ∠DAB = ∠CBA

(i) In ΔABD and ΔBAC,

AB = BA (Common)

∠DAB = ∠CBA (Given)

AD = BC (Given)

Therefore, ΔABD ≅ ΔBAC by SAS congruence condition.

(ii) Since, ΔABD ≅ ΔBAC

Therefore BD = AC by CPCT

(iii) Since, ΔABD ≅ ΔBAC

Therefore ∠ABD = ∠BAC by CPCT

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

**Answer**

Given,

AD and BC are equal perpendiculars to AB.

To prove,

CD bisects AB

Proof,

In ΔAOD and ΔBOC,

∠A = ∠B (Perpendicular)

∠AOD = ∠BOC (Vertically opposite angles)

AD = BC (Given)

Therefore, ΔAOD ≅ ΔBOC 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 (see Fig. 7.19). Show that ΔABC ≅ ΔCDA.**

**Answer**

Given,

l || m and p || q

To prove,

ΔABC ≅ ΔCDA

Proof,

In ΔABC and ΔCDA,

∠BCA = ∠DAC (Alternate interior angles)

AC = CA (Common)

∠BAC = ∠DCA (Alternate interior angles)

Therefore, ΔABC ≅ ΔCDA by ASA congruence condition. **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.**

Answer

Given,

l is the bisector of an angle ∠A.

BP and BQ are perpendiculars.

(i) In ΔAPB and ΔAQB,

∠P = ∠Q (Right angles)

∠BAP = ∠BAQ (l is bisector)

AB = AB (Common)

Therefore, ΔAPB ≅ ΔAQB by AAS congruence condition.

(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of ∠A.**6. In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.**

**Answer**

Given,

AC = AE, AB = AD and ∠BAD = ∠EAC

To show,

BC = DE

Proof,

∠BAD = ∠EAC (Adding ∠DAC both sides)

∠BAD + ∠DAC = ∠EAC + ∠DAC

⇒ ∠BAC = ∠EAD

In ΔABC and ΔADE,

AC = AE (Given)

∠BAC = ∠EAD

AB = AD (Given)

Therefore, ΔABC ≅ ΔADE 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 ∠BAD = ∠ABE and ∠EPA = ∠DPB (see Fig. 7.22). Show that (i) ΔDAP ≅ ΔEBP (ii) AD = BE**

**Answer**

Given,

P is mid-point of AB.

∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) ∠EPA = ∠DPB (Adding ∠DPE both sides)

∠EPA + ∠DPE = ∠DPB + ∠DPE

⇒ ∠DPA = ∠EPB

In ΔDAP ≅ ΔEBP,

∠DPA = ∠EPB

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

∠BAD = ∠ABE (Given)

Therefore, ΔDAP ≅ ΔEBP by ASA congruence condition.

(ii) AD = BE by CPCT.

**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 = 1/2 AB**

**Answer**

Given,

∠C = 90°, M is the mid-point of AB and DM = CM

(i) In ΔAMC and ΔBMD,

AM = BM (M is the mid-point)

∠CMA = ∠DMB (Vertically opposite angles)

CM = DM (Given)

Therefore, ΔAMC ≅ ΔBMD by SAS congruence condition.

(ii) ∠ACM = ∠BDM (by CPCT)

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

Now,

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

⇒ 90° + ∠B = 180°

⇒ ∠DBC = 90°

(iii) In ΔDBC and ΔACB,

BC = CB (Common)

∠ACB = ∠DBC (Right angles)

DB = AC (byy CPCT, already proved)

Therefore, ΔDBC ≅ ΔACB by SAS congruence condition.

(iv) DC = AB (ΔDBC ≅ ΔACB)

⇒ DM = CM = AM = BM (M is mid-point)

⇒ DM + CM = AM + BM

⇒ CM + CM = AB

⇒ CM = 1/2AB**Exercise 7.2****1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that : (i) OB = OC **

**(ii) AO bisects ∠A**

**Answer**

Given,

AB = AC, the bisectors of ∠B and ∠C intersect each other at O

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

∴ ∠B = ∠C

⇒ 1/2∠B = 1/2∠C

⇒ ∠OBC = ∠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.

∠BAO = ∠CAO (by CPCT)

Thus, AO bisects ∠A.**2. In ΔABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ΔABC is an isosceles triangle in which AB = AC.**

**Answer**

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 (by CPCT)**3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.**

**Answer**

Given,

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 (see Fig. 7.32). Show that (i) ΔABE ≅ ΔACF (ii) AB = AC, i.e., ABC is an isosceles triangle.**

**Answer**

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 (see Fig. 7.33). Show that ∠ABD = ∠ACD.**

**Answer**

Given,

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 (see Fig. 7.34). Show that ∠BCD is a right angle.**

**Answer**

Given,

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.****Answer**

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.**

**Answer**

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.

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