Question

In the given figure, ABC is a triangle right angled at B. D and E are points on BC trisect it.
Prove that 8AE² = 3AC² + 5AD².

Answer 1

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Answer 2

Answer:

Given information: ABC is a triangle right angled at B. D and E are points on BC trisect it.

BD=DE=EC

To prove: 8AE² = 3AC² + 5AD²

Proof:

According to the Pythagoras theorem, in a right angled triangle

$hypotenuse^2=base^2+perpendicular^2$

Using Pythagoras theorem we get

$AD^2=AB^2+BD^2$           .... (1)

$AE^2=AB^2+BE^2$

$AE^2=AB^2+(2BD)^2$               (BE=2BD)

$AE^2=AB^2+4BD^2$            .... (2)

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

$AC^2=AB^2+(3BD)^2$                (BC=3BD)

$AC^2=AB^2+9BD^2$              .... (3)

Taking RHS,

$RHS=3(AC^2)+5(AD^2)$

Using (1) and (3) we get

$RHS=3(AB^2+9BD^2)+5(AB^2+BD^2)$

$RHS=3AB^2+27BD^2+5AB^2+5BD^2$

$RHS=8AB^2+32BD^2$

$RHS=8(AB^2+4BD^2)$

Using equation (2) we get

$RHS=8(AE^2)$

$RHS=LHS$

Hence proved LHS=RHS.