ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2 AD.
Solution:
Given, ABC is a right triangle with AB = AC
Bisector of angle A meets BC at D.
We have to prove that BC = 2 AD.
In triangle ABC,
Given, AB= AC
We know that angles opposite to equal sides are equal.
So, ∠C = ∠B ------------------------- (1)
∠4 = ∠3
∠A + ∠B + ∠C = 180°
From (1),
90° + ∠B + ∠B = 180°
2 ∠B = 180° - 90°
2 ∠B = 90°
∠B = 90°/2
∠B = 45°
We know ∠B = ∠C
So, ∠C = 45°
i.e., ∠3 = ∠4 = 45°
Also, ∠1 = ∠2 = 45°
This implies AD is the bisector of angle A.
We observe that ∠1 = ∠3 and ∠2 = ∠4
We know that the sides opposite to equal angles are equal.
BD = AD
DC = AD
From the figure,
BC = BD + CD
BC = AD + AD
BC = 2AD
Therefore, BC = 2AD
✦ Try This: In a triangle ABC, AB=AC and ∠A=36°. If the internal bisector of ∠C meets AB at a point D, prove that AD=BC.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 6
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2 AD
Summary:
The (interior) bisector of an angle, also called the internal angle bisector, is the line or line segment that divides the angle into two equal parts. ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. It is proven that BC = 2 AD
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