निम्नलिखित आकृति में, $\mathrm{AD} \perp \mathrm{BC}$ और $\frac{\mathrm{BD}}{\mathrm{DA}}=\frac{\mathrm{DA}}{\mathrm {DC}}$ फिर $\angle BAC=?$

$\Delta \mathrm{s} \mathrm{BDA}$ और $\Delta \mathrm{ADC}$ में,

$\frac{\mathrm{DB}}{\mathrm{DA}}=\frac{\mathrm{DA}}{\mathrm{DC}}$

और $\angle B D A=\angle A D C$

$\therefore$ By SAS मापदंड

$\triangle \mathrm{BDA} \sim \triangle \mathrm{ADC}$

$\Rightarrow \angle \mathrm{ABD}=\angle \mathrm{CAD}$ and $\angle \mathrm{BAD}=\angle \mathrm{ACD}$

$\Rightarrow \angle \mathrm{ABD}+\angle \mathrm{ACD}=\angle \mathrm{CAD}+\angle \mathrm{BAD}$

$\Rightarrow \angle \mathrm{B}+\angle \mathrm{C}=\angle \mathrm{A}$

$\Rightarrow \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=2 \angle \mathrm{A}$

$\Rightarrow 2 \angle \mathrm{A}=180^{\circ} \Rightarrow \angle \mathrm{A}=90^{\circ}$