Question

in triangle ABC, given that AB=AC and BD perpendicular to AC. Prove that BC(BC)= 2AC•CD

Answer 1

solution of your Qus.

Answer 2

Explanation:

The image of the question is attached below.

ΔABD and ΔBCD are right triangles.

In ΔABD,

Using Pythagoras theorem,

$A B^{2}=A D^{2}+B D^2$

Subtract AD² on both sides of the equation.

$A B^{2}-A D^{2}=B D^2$

Switch the sides

$B D^2=A B^{2}-A D^{2}$ – – – – (1)

In ΔBCD,

Using Pythagoras theorem,

$B C^{2}=C D^{2}+BD^2$

Subtract CD² on both sides of the equation.

$B C^{2}-C D^{2}=B D^{2}$

Switch the sides

$B D^{2}=B C^{2}-C D^{2}$ – – – – (2)

Compare equation (1) and equation (2),

$A B^{2}-A D^{2}=B C^{2}-C D^{2}$

$A B^{2}+C D^{2}=B C^{2}+A D^{2}$

$A C^{2}+C D^{2}=B C^{2}+(AC-CD)^{2}$

Using algebraic identity: $(a-b)^2=a^2-2ab+b^2$

$A C^{2}+C D^{2}=B C^{2}+AC^2-2 AC\cdot CD+CD^2$

Subtract $AC^2+CD^2$ from both side of the equation.

$0=B C^{2}-2 AC\cdot CD$

$2 AC\cdot CD=B C^{2}$

$B C^{2}=2 AC\cdot CD$

BC(BC) = 2 AC. CD

Hence proved.

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