Question

In $△ABC$, B is a right angle, and BD is perpendicular to AC and DE is perpendicular to BC. Then which of the following is true?

• A. DC² × ED = EC² × AD
• B. ED² × DC = EC² × AD
• C. ED² × DC = AD² × EC
• D. DC² × ED = AD² × EC

Answer 1

The correct option is B

ED² × DC = EC² × AD

In $△ABC$ and $△BDC$

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

$\angle C=\angle C$ (common angle)

Therefore, $△ABC\sim △BDC$ by AA similarity

$\frac{BC}{DC}=\frac{AC}{BC}\Rightarrow {BC}^2=AC\times DC$ ------------------------------------ I

Similarly $△BDC\sim △DEC$

Therefore $\frac{DC}{EC}=\frac{BC}{DC}\Rightarrow {DC}^2=EC\times BC$

$BC=\frac{{DC}^2}{EC}$ ------------------------------------------------------------------------------II

Substitute II in I

$\frac{{DC}^4}{{EC}^2=AC\times DC}$

${DC}^3={EC}^2\times AC$

${DC}^2\times DC={EC}^2\times (AD+DC)$

$({EC}^2+{ED}^2)\times DC={EC}^2(AD+DC)({DC}^2={EC}^2+{ED}^2)$

Therefore, ${ED}^2\times DC={EC}^2\times AD$