Question

In the adjoining figure, ABC is a right-angled triangle, right-angled at A such that AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.

Answer 1

Given: Triangle ABC is right-angled at A and AB = AC. The bisector of ∠A meets BC at D.

To Prove: BC = 2AD

Proof (Step-wise):

1. Let AB = AC = s (length s > 0).

Since ∠A = 90° and AB = AC, triangle ABC is an isosceles right triangle, so the base angles ∠B and ∠C are each 45°.

2. Place the triangle on a coordinate plane for a clear calculation:

Put A at the origin A(0, 0).

AB along the x-axis and AC along the y-axis. 

Then B(s, 0) and C(0, s).

3. The angle bisector of ∠A which is 90° is the line at 45° through A, i.e., the line y = x. 

The line BC has equation x + y = s, the line through (s, 0) and (0, s).

4. The intersection D of y = x and x + y = s satisfies 2x = s.

So, `x = s/2`. 

Hence, `D = (s/2, s/2)`.

5. Compute lengths:

BC = Distance between B(s, 0) and C(0, s)

= `sqrt((s - 0)^2 + (0 - s)^2)` 

= `s xx sqrt(2)`

AD = Distance from A(0, 0) to `D(s/2, s/2)`

= `sqrt((s/2)^2 + (s/2)^2)`

= `(s/2) xx sqrt(2)`

6. Compare:

`BC = s xx sqrt(2)`

= `2 xx (s/2) xx sqrt(2)` 

= 2 × AD

Therefore, BC = 2AD, as required.