Question

If ABC is a right angled triangle such that AB = AC and bisector of
angle C intersects the side AB at D, then prove that AC + AD = BC.​

Answer 1

Answer:

let AC = x

then BC =  2–√ x

as AD and DB will be in ratio of AC and BC by angle bisector theorem

so let AD = y and so DB =  2–√ y

but AD + DB = AB = AC

so y+ 2–√ y = x

y( 2–√  + 1) = x

so y =  x2–√+1  

on rationalising the denominator it results in

y = ( 2–√  - 1) x

add x on both sides

y + x= ( 2–√  - 1) x + x

AD + AC = ( 2–√  - 1 + 1) x

AD + AC = ( 2–√ ) x

so AC + AD = BC

Step-by-step explanation:

Answer 2

Answer:

AC +AD = m + (√2–1)m = √2m = BC.

• Proved.

• Originally Answered: ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. ... In triangle ABC, side AC and the perpendicular bisector of BC meet in point D, and BD bisects angle ABC.