Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

In Fig., $\angle BAC=90^{o}$ and $AD\perp BC$ . Then,

(a) $BD.CD=BC^{2}$

(b) $AB.AC=BC^{2}$

(c) $BD.CD=AD^{2}$

(d) $AB.AC=AD^{2}$

Answers (1)

Given :- $\angle BAC=90^{o}$

First of all we have find $\angle DBA$ and $\angle DAC$ in triangles $ABD$ and $ADC$ respectively

In $\Delta ABD$ and $\angle ADC$

$\angle ADB=90^{o}$

$\angle ADC=90^{o}$

In $\Delta ABC$

$\angle BAC=90^{o}$ (given)

Now let us find angle CAD from triangle ADC

Here $\angle ADC+\angle DCA+\angle CAD=180^{o}$

(sum of interior angles of triangle = 180^o)

$90^{o}+\angle C+\angle CAD=180^{o}$

$\angle CAD=180^{o}-90^{o}-\angle C$

$\angle CAD=90^{o}-\angle C \; \; \; \; \; \; \; \; \; \; \; \; \; ....(1)$

Now let us find angle DBA using triangle ABC

Here $\angle ABC+\angle BCA+\angle CAB=180^{o}$

{sum of interior angles of triangle = 180^o}

$\angle ABC+\angle C+90^{o}=180^{o}$

$\angle ABC=90^{o}-\angle C$

$\angle ABC=\angle DBA=90^{o}-\angle C \; \; \; \; \; \; \; ...(2)$

In $\Delta ABD$ and $\Delta ADC$ we get

$\angle ADB=\angle ADC$ {90^o each}

$AD=AD$ {common side}

$\angle CAD=\angle DBA$ {from equation (1) and (2) it is clear that both is $90^{o}-\angle C$ }

$\therefore \Delta ABD\sim \Delta ADC$ (by ASA similarity)

$\therefore \frac{BD}{AD}=\frac{AD}{CD}$

$BD.CD=AD.AD$ {by cross multiplication}

$BD.CD=AD^{2}$

Hence option (C) is correct.

Posted by

infoexpert23

View full answer

Similar Questions

Latest Question