Question

3. In the figure 10.23, ABC is a triangle and AD is an altitude. show that---
1. BP || AD
2. CQ || AD
3. BP || CQ


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Answer 1

GIVEN :-

• ABC is a triangle.
• AD is an altitude.

TO PROVE :-

• BP || AD.
• CQ || AD.
• BP || CQ.

PROOF :-

✭ BP || AD.

➟ ∠BAD = ∠PBA = 40°. [Alternate interior angles ]

As we know that , alternate interior angles is one of the property of a parallel lines.

∴ BP || AD.

✯ CQ || AD.

➙ ∠ADC = ∠DCQ = 90°. [Alternate interior angles ]

As we know that , alternate interior angles is one of the property of a parallel lines.

∴ CQ || AD.

✪ BP || CQ.

Here in ∆ ADB , By angle sum property of triangle.

➟ ∠BAD + ∠ABD + ∠ADB = 180°

➟ 40° + ∠ABD + 90° = 180°

➟ ∠ABD + 130° = 180°

➟ ∠ABD = 180° - 130°

➟ ∠ABD = 50°

Now,

➙ ∠PBD = ∠PBA + ∠ABD

➙ ∠PBD = 40° + 50°

➙ ∠PBD = 90°

Therefore,

➙ ∠PBD = ∠DCQ = 90° . [Alternate interior angles ]

As we know that , alternate interior angles is one of the property of a parallel lines.

∴ BP || CQ.

Answer 2

Question:-

In the figure 10.23, ABC is a triangle and AD is an altitude. show that---

1. BP || AD

2. CQ || AD

3. BP || CQ

Given:-

∆ABC

∠BAD = 40°

∠ABP = 40°

D = 90°

C = 90°

To prove :- i. BP ll AD

ii. AD ll CQ

iii. BP II CQ

Proof:-

∠BAD = ∠ABP [ 40° ]

Since, BAD and ABP are alternative interior angles and are equal.

∴ BP II AD (ii)

And ∠ADC = DC

Same as in equation (ii)

AD = CQ (iii)

Now fom equation (ii) and (iii)

BP II CQ

Hence Verified