Question

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

• 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, $△ADB\sim △BDCbyAAsimilarity$
$\frac{AD}{DB}=\frac{BD}{DC}=\frac{AB}{BC}$
$\Rightarrow B{D}^2=AD.DC$

$\angle DBC=\angle DBE$ (Common angle)
$\angle BDC=\angle DEB={90}^{\circ }$
Similarly $△BDC\sim △BED$

$\Rightarrow$ $\frac{BD}{BE}=\frac{DC}{ED}=\frac{BC}{BD}$
$\Rightarrow B{D}^2=BE\times BC\Rightarrow B{D}^2=AD\times DC=BE\times BC$