Question

In $\Delta$ABC, B is a right angle, and BD is perpendicular to AC and DE is perpendicular to BC. Find the value of AD.DC .

Answer 1

In $\Delta ABD\&\Delta BDC\angle ADB=\angle BDC={90}^{\circ }\angle A=\angle DBC$ ( Since in $\Delta BDC,\angle DBC={90}^{\circ }-\angle C=\angle A$
In $\Delta ABC\angle A={90}^{\circ }-\angle C)\therefore ,\Delta ADB\sim \Delta BDC$ by AA-similarity
$\frac{AD}{DB}=\frac{BD}{DC}=\frac{AB}{BC}\Rightarrow B{D}^2=AD.DC$
Similarly $\Delta BDC\sim \Delta BED\angle DBC=\angle DBE\left(\text{Common angle}\right)\angle BDC=\angle DEB={90}^{\circ }\frac{BD}{BE}=\frac{DC}{ED}=\frac{BC}{BD}\Rightarrow B{D}^2=BE.BC\Rightarrow B{D}^2=AD.DC=BE.BC$