Question

In triangle ABC, $\angle$ A = ${90}^{\circ },$ CA = AB and D is a point on AB produced. $D{C}^2-B{D}^2=$

• A. $2AB.AC$
• B. $2AB.AD$
• C. $2A{D}^2$
• D. $2DC.AD$

Answer 1

The correct option is B $2AB.AD$

Given: $△ABC$, $\angle A={90}^{\circ }$, $AC=AB$
Now, in $△ADC$
$D{C}^2=A{C}^2+A{D}^2$
Thus, $D{C}^2-B{D}^2=A{C}^2+A{D}^2-B{D}^2$
$D{C}^2-B{D}^2=A{C}^2+(AB+BD{)}^2-B{D}^2$
$D{C}^2-B{D}^2=A{C}^2+A{B}^2+B{D}^2+2AB.BD-B{D}^2$
$D{C}^2-B{D}^2=2A{B}^2+2AB.BD$ (Since, $AB=AC$)
$D{C}^2-B{D}^2=2AB(AB+BD)$
$D{C}^2-B{D}^2=2AB.AD$