Question

$ABD$ is a triangle right angled at $A$ and $AC\perp BD$. Find $x$.

• A. $CA$
• B. $BD$
• C. $CB$

Answer 1

Perpendicular from right angle to hypotenuse divides the triangle into two similar triangles and these are similar to the whole triangle also.

$\therefore \Delta ABD\sim \Delta CBA---\left(i\right)\Delta ABD\sim \Delta CAD---\left(ii\right)\Delta CBA\sim \Delta CAD---\left(iii\right)$

From (i)
$\frac{AB}{CB}=\frac{AD}{CA}\to x=CA$

From (ii)
$\frac{AB}{CA}=\frac{BD}{AD}\to x=BD$

From (iii)
$\frac{CB}{CA}=\frac{BA}{AD}\to x=CB$