Question

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

• A. ${DC}^2\times ED={EC}^2\times AD$
• B. $E{D}^2\times DC=E{C}^2\times AD$
• C. $E{D}^2\times DC=A{D}^2\times EC$
• D. $D{C}^2\times ED=A{D}^2\times EC$

Answer 1

The correct option is B

$E{D}^2\times DC=E{C}^2\times AD$

In $△ABC$ and $△BDC$

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

$\angle BCA=\angle ECD$ (common in both the triangles)

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

$\frac{BC}{DC}=\frac{AC}{BC}\Rightarrow {BC}^2=AC\times DC....\left(1\right)$

Similarly, $△BDC\sim △DEC$

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

$\Rightarrow BC=\frac{{DC}^2}{EC}....\left(2\right)$

On substituting (2) in (1), we get

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

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

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

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

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