Question

In figure, $\angle BAC={90}^{\circ }$ and segment $AD\perp BC$. Prove that ${AD}^2=BD\times DC$. [3 MARKS]

Answer 1

Concept: 1 Mark
Proof : 2 Marks

In $\Delta ABC$,



$\angle BAC={90}^{\circ }$ [Given]

Let $\angle ABC=x^{\circ }and\angle ACB=(90-x{)}^{\circ }.....(1)$ [Since they are complementary]

In $\Delta ABD$,

$\angle ADB={90}^{\circ }$ [ $\because AD\perp BC$]

$\angle ABD=\angle ABC=x^{\circ }$ [From (1)]

$\angle BAD=(90-x{)}^{\circ }$ [ Complemntary angle]

SImilarly, In $\Delta ACD$,

$\angle ADC={90}^{\circ },\angle ACD=\angle ACB=(90-x{)}^{\circ },\angle DAC=x^{\circ }$

Now, In $\Delta ABD$ and $\Delta ACD$,

$\angle ADB=\angle ADC$ [Each equal to ${90}^{\circ }$]

$\angle DBA=\angle DAC$ [Each equal to $x$]

$\therefore \Delta DBA\sim \Delta DAC$ [AA - criterion of similarity]

$\Rightarrow \frac{DB}{DA}=\frac{DA}{DC}$ [In similar triangles corresponding sides are proportional]

$\Rightarrow {AD}^2=BD\times DC$