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$\cdot$DC. $\left[3Marks\right]$

Answer 1

In $\Delta ABD\&\Delta BDC\angle ADB=\angle BDC={90}^{\circ }\angle A=\angle DBC(\because \text{in}\Delta BDC,\angle DBC={90}^{\circ }-\angle C\text{and in}\Delta ABC,\angle A={90}^{\circ }-\angle C)$
$\therefore \Delta ADB\sim \Delta BDC,\text{by AA-similarity}$
$\left[1Mark\right]$
$\therefore \frac{AD}{DB}=\frac{BD}{DC}=\frac{AB}{BC}$
$\Rightarrow B{D}^2=AD\cdot DC$
$\left[1Mark\right]$

Similarly,
$\Delta BDC\sim \Delta BED\angle DBC=\angle DBE\left(\text{Common angle}\right)\angle BDC=\angle DEB={90}^{\circ }\therefore \frac{BD}{BE}=\frac{DC}{ED}=\frac{BC}{BD}$
$\Rightarrow B{D}^2=BE\cdot BC$
$\left[0.5Marks\right]$
$\Rightarrow AD\cdot DC=BE\cdot BC$
$\left[0.5Marks\right]$