108
In $\triangle$ABC, if AD $\perp$ BC and AD$^2$ = BD $\times$ DC, then prove that $\angle$BAC = $90^{\circ}$.
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Sol.
Correct figure
AD$^2$ = BD $\times$ DC
$\frac{AD}{DC} = \frac{BD}{AD}$
$\triangle$ADB $\sim \triangle$CDA
$\angle$BAD = $\angle$ACD & $\angle$DAC = $\angle$DBA
In $\triangle$ABC
$\angle$ABC + $\angle$CAB + $\angle$BCA = $180^{\circ}$
$\angle$ABC + $\angle$BAD + $\angle$CAD + $\angle$BCA = $180^{\circ}$
$\angle$BAD + $\angle$CAD = $90^{\circ}$
$\angle$BAC = $90^{\circ}$
Correct figure
AD$^2$ = BD $\times$ DC
$\frac{AD}{DC} = \frac{BD}{AD}$
$\triangle$ADB $\sim \triangle$CDA
$\angle$BAD = $\angle$ACD & $\angle$DAC = $\angle$DBA
In $\triangle$ABC
$\angle$ABC + $\angle$CAB + $\angle$BCA = $180^{\circ}$
$\angle$ABC + $\angle$BAD + $\angle$CAD + $\angle$BCA = $180^{\circ}$
$\angle$BAD + $\angle$CAD = $90^{\circ}$
$\angle$BAC = $90^{\circ}$