Question

ABC is a right triangle such that AB = AC and bisector or angle C intersects the side AB
at D. Prove that AC + AD = BC.
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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

We have an equilateral right angle triangle,

Let AB=AC=1

the hypotenuse BC= sqrt(2) =r2

Applying the sine rule in the divided triangles gives

AD/1 =( 1-AD)/r2

Hence

AD (1+r2)=1

AD (1+r2)(r2-1)=r2-1

AD = r2-1

Or

r2 =1 +AD

BC=AC+AD

Step-by-step explanation: