Question

In $△$ABC, $\angle B$ is a right angle, and BD is perpendicular to AC and DE is perpendicular to BC. Then $AD.DC=$

• A. $B{D}^2$
• B. $BE.BC$
• C. $AB.BC$
• D. $BD.DC$

Answer 1

Answer: (A), (B)

The correct options are
A

$B{D}^2$

B

$BE.BC$

In $△ABD$ and $△BDC$,

$\angle ADB=\angle BDC={90}^{\circ }$

$\angle A=\angle DBC$

(Since, in $△BDC,\angle DBC={90}^{\circ }-\angle C=\angle A$ and in $△ABC,\angle A={90}^{\circ }-\angle C$)

Therefore, $△ADB\sim △BDC$ (by AA similarity criterion)

$\Rightarrow \frac{AD}{DB}=\frac{BD}{DC}=\frac{AB}{BC}$

$\Rightarrow B{D}^2=AD.DC$

Also, in $△BDC$ and $△BED$ ,

$\angle DBC=\angle DBE$ [Common angle]

$\angle BDC=\angle BED$ [${90}^{\circ }$]

Thus, by AA similarity criterion, we have $△BDC\sim △BED$

$\Rightarrow \frac{BD}{BE}=\frac{DC}{ED}=\frac{BC}{BD}$

$\Rightarrow B{D}^2=BE.BC$

$\therefore B{D}^2=AD.DC=BE.BC$