Question

In right triangle ABC angle C =90° and D, E, F are three points on BC (B, D, E, F, C) such that thet divide it in equal parts. Then prove that 8 (AF^2 + AD^2) = 11AC^2 + 5 AB^2​

Answer 1

the required answer in in the attachment..

Answer 2

Given: In ΔABC, ∠C = 90°

D, E, F are three points on BC such that they divided it into equal parts.

To prove: $8(AF^2+AD^2)=11AC^2+5AB^2$

Proof:

D, E, F are three points on BC such that they divided it into equal parts. Therefore, BD = DE = EF = FC

$DC=\dfrac{3}{4}BC$

$FC=\dfrac{1}{4}BC$

• In ΔAFC, ∠C = 90° , Using Pythagoras theorem

$AF^2=AC^2+FC^2$

$AF^2=AC^2+\dfrac{BC^2}{16}$ ---- (1)              $\because FC=\dfrac{1}{4}BC$

• In ΔADC, ∠C = 90° , Using Pythagoras theorem

$AD^2=AC^2+DC^2$

$AD^2=AC^2+\dfrac{9BC^2}{16}$ -----(2)              $\because FC=\dfrac{3}{4}BC$

Add eq(1) and eq(2)

$AF^2+AD^2=AC^2+\dfrac{BC^2}{16}+AC^2+\dfrac{9BC^2}{16}$

$AF^2+AD^2=2AC^2+\dfrac{10BC^2}{16}$

$AF^2+AD^2=\dfrac{32AC^2+10BC^2}{16}$

$AF^2+AD^2=\dfrac{32AC^2+10AB^2-10AC^2}{16}$      $\because AB^2=AC^2+BC^2$

$AF^2+AD^2=\dfrac{22AC^2+10AB^2}{16}$

$8(AF^2+AD^2)=11AC^2+5AB^2$

hence proved