Question

In the given figure, $△ABC$ is a right angled triangle and $BD$ is perpendicular to $AC$. Which of the following is correct?

• A. $△ABC\sim △ADB$
• B. $△ADB\sim △BDC$
• C. $△ABC\sim △BDC$
• D. All of the above.

Answer 1

The correct option is D All of the above.

The given figure is as follows:

In $△ABC$ and $△ADB,$
$\angle A=\angle A$ (common angle)
$\angle ABC=\angle ADB$ (both equal to ${90}^o$)
So, $△ABC\sim △ADB$ (AA property)

In $△ABC$ and $△BDC$
$\angle C=\angle C$ (common angle)
$\angle ABC=\angle BDC$ (both equal ${90}^o$)
So, $△ABC\sim △BDC$ (AA property)

As $△ABC$ is similar to both triangles,
$\therefore △ADB\sim △BDC$

Hence, all are correct.
So, option $\left(d\right)$ correct.