Question

In $△BAC,\angle A={90}^0.$ if $AD\perp BC$ Then prove that $△BDA$~ $△ADC.$

Answer 1

$⟹let\angle ACD=\theta$

$In△ABC:$

$\angle A+\angle B+\angle C={180}^{\circ }$

$⟹{90}^{\circ }+\angle B+\theta ={180}^{\circ }$

$⟹\angle B=90-\theta -\left(1\right)$

$In△ADC$

$⟹\angle DAC+\angle ADC+\angle ACD={180}^{\circ }$

$⟹\angle DAC+{90}^{\circ }+\theta ={180}^{\circ }$

$⟹\angle DAC=90-\theta -\left(2\right)$

$In△BDA\&△ADC$

$⟹\angle BDA=\angle ADC={90}^{\circ }$

$\angle DBA=\angle DAC-\left(from\right(1\left)\&\right(2\left)\right)$

$\therefore △BDA\sim △ADCbyA.Asimilarity.$