Prove that ac2 = ap2 + 2(1+2)bp2, Mathematics

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ABC is a right-angled isosceles triangle, right-angled at B. AP, the bisector of ∠BAC, intersects BC at P. Prove that AC² = AP² + 2(1+√2)BP²

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Ans: AC = √2AB (Since AB = BC)

AB/AC = BP/CP (Bisector Theorem)

⇒ CP = √2 BP

AC² - AP² = AC² - (AB² + BP²)

= AC² - AB² - BP²

= BC² - BP²

= (BP + PC)² - BP²

= (BP + √2BP)² - BP²

= 2BP² + 2√2 BP²

= 2 ( √2 +1) BP² ⇒ AC² = AP² + 2(1+√2)BP²

Proved

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