**Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that**

**(i) Δ ABD ≅ Δ ACD**

**(ii) Δ ABP ≅ Δ ACP**

**(iii) AP bisects ∠ A as well as ∠ D.**

**(iv) AP is the perpendicular bisector of BC.**

Solution:

(i)In ∆ABD and ∆ACD ,we have,

AB = AC *[given]*

BD = CD *[given]*

AD = AD *[common]*

*⸫* ∆ABD ≅ ∆ACD *[ SSS congruence]*

(ii) In ∆ABP and ∆ACP ,we have,

AB = AC *[given]*

∠BAP = ∠CAP *[given]*

AP = AP *[common]*

*⸫* ∆ABP ≅ ∆ACP *[ SSS congruence]*

(iii) ∆ABD ≅ ∆ADC [From part (i)]

⇒ ∠ADB = ∠ADC (CPCT)

⇒180 ° – ∠ADB = 180° – ∠ADC

⇒Also, from part (ii), ∠BAPD = ∠CAP (CPCT)

*⸫*AP bisects DA as well as ∠D

(iv) ∆BDP = ∆CDP (which is already proved)

BP = CP (by CPCT)

But we have, ∠BPA + ∠CPA = 180 °

2 ∠BPA = 180 °

∠BPA = 90 °

Therefore, AP is perpendicular bisector of BC.

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

**(i) AD bisects BC**

**(ii) AD bisects ∠ A.**

Solution:

In right angled triangle ADB and right angled triangle ADC,

AB = AC (given)

AD = AD (common)

** ⸫**∆ADB ≅ ∆ADC (RHS rule)

** ⸫**BD = CD (CPCT)

⇒AD bisects BC

(ii) ∆ADB≅∆ADC (which is already proved)

*⸫*∠BAD = ∠CAD (CPCT)

⇒AD bisects ∠A

**Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR (see Fig. 7.40). Show that:**

**(i) Δ ABM ≅ Δ PQN**

**(ii) Δ ABC ≅ Δ PQR**

Solution:

Given, Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR

We have to prove,

(i) Δ ABM ≅ Δ PQN

(ii) Δ ABC ≅ Δ PQR

Let us prove (i), In ∆ABM and ∆PQN ……..(1)

AB = PQ …………(2) (given)

AM = PN (given)

BC = QR (M and N are the midpoint of BC and QR respectively)

2BM = 2QN

BM = QN ………..(3)

From (1) (2) and (3), we have,

Δ ABM ≅ Δ PQN (SSS rule)

(ii) ∆ABM ≅ ∆PQN (which is already proved)

⇒ ∠ABM = ∠PQN (CPCT)

⇒ ∠ABC = ∠PQR ……….(4)

In ∆ABC and ∆PQR,

AB = PQ (given)

BC = QR (given)

∠ABC = ∠PQR (from (4))

*⸫* ∆ABC ≅ ∆PQR

**BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.**

Solution:

BE and CF are two equal altitudes of a triangle ABC

We have to prove, ∆ABC is an isosceles triangle

In right angled triangle BFC and right angled triangle CFB,

BE = CF (given)

BC = CB (common)

** ⸫** ∆BEC ∆CFB (RHS rule)

∠BCE = ∠CBF (CPCT)

AB = AC (sides opposite to equal angles of a triangle are equal)

∆ABC is an isosceles triangle.

**ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.**

Solution:

ABC is an isosceles triangle with AB = AC

We have to prove ∠B = ∠C

Draw a line AP⊥BC

In right angled triangle APB and right angled triangle APC,

AB = AC (given)

AP = AP (common)

*⸫*∆APB≅∆APC (RHS rule)

∠ABP = ∠ACP (CPCT)

⇒∠B = ∠C

## 1 thought on “Triangles – Exercise 7.3 – Class IX”

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