Question

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

• A. $△ABD\sim △BDC$
• B. $△ABD\sim △ABC$
• C. $△BCD\sim △ABC$
• D. $△BDC\sim △ABC$

Answer 1

The correct option is D $△BDC\sim △ABC$

Let $\angle A=x$
We knoe that sum of interior angles of a triangle $={180}^{\circ }$
So, $\angle A+\angle B+\angle C={180}^{\circ }$
$\angle C={180}^{\circ }-({90}^{\circ }+x)=({90}^{\circ }-x)$
Similarly $\angle ABD={180}^{\circ }-({90}^{\circ }+x)=({90}^{\circ }-x)$
$\angle CBD={90}^{\circ }-({90}^{\circ }-x)=x$
In $△BDC$ and $△ABC$
$\angle BDC=\angle ABC\left(each{90}^{\circ }\right)$

$\angle A=\angle CBD$
By using AA criteria,
$△BDC\sim △ABC$