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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
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
1+2+3+78+980
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