Question

Triangle ABC is a right triangle right angled at A such that AB =AC and bisector of angle C intersects the side AB at D. Prove that AC +AD =BC

Answer 1

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✔️Let AB = AC = a and AD = b

✔️In a right angled triangle ABC , BC2 = AB2 + AC2

BC2 = a2 + a2

BC = a√2

✔️Given AD = b, we get.....

DB = AB – AD or DB = a – b

✔️We have to prove that AC + AD = BC or (a + b) = a√2.

✔️By the angle bisector theorem, we get.....

AD/ DB = AC / BC

b/(a - b) = a/ a√2

b/(a - b) = 1/√2

b = (a – b)/ √2

b√2 = a – b

b(1 + √2) = a

b = a/ (1 + √2)

✔️Rationalizing the denominator with (1 - √2)

b = a(1 - √2) / (1 + √2) × (1 - √2)

b = a(1 - √2)/ (-1)

b = a(√2 - 1)

b = a√2 – a

b + a = a√2

or  AD + AC = BC [we know that AC = a, AD = b and BC = a√2]

✔️Hence it is proved.

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Answer 2

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