Question

ABC is an isosceles triangle right angled at B. AP the bisector of /_ BAC,intersects BC at P. Prove that AC^2 = AP^2 +2(1+root 2 ) BP ^2​

Answer 1

Answer:

Please see the attachment

Answer 2

GIVEN :

ABC ,an isosceles triangle right angled at B.

AP the bisector of /_ BAC,intersects BC at P.

TO FIND :

Prove that AC^2 = AP^2 +2(1+√2 ) BP ^2​

SOLUTION :

◆Since it's an right angle isosceles triangle,

<BAC = 45°

AB = BC

◆Since AP bisects BC ,

< BAP = 45/2 =22.5

◆AC^2 = AP^2 +2(1+√2 ) BP ^2​

◆Considering RHS ,

AP^2 +2(1+√2 ) BP ^2​

= AB^2 + PB^2 + 2(1+√2) PB^2

◆[By, Pythagoras theorem

AP^2 = AB^2 + PB^2]

=AB^2 + (3 + 2√2)PB^2

= AB^2 + (3 + 2√2)(√2 -1) AB^2

◆[Tan 22.5 = (√2-1) = PB/AB

PB = (√2-1)AB]

= AB^2 + (9-8)AB^2

=AB^2 + BC^2 = AC^2

=LHS

Hence the solution.