ABC is a right angled triangle with $\angle \mathrm{ABC}=90^{\circ} . \mathrm{D}$ is any point on AB and DE is perpendicular to AC , Prove that:
(a) $\triangle \mathrm{ADE} \sim \triangle \mathrm{ACB}$
(b) If $\mathrm{AC}=13 \mathrm{~cm}, \mathrm{BC}=5 \mathrm{~cm}$ and $\mathrm{AE}=4 \mathrm{~cm}$, find DE and AD .

View full solution →(a) $\triangle \mathrm{ADE} \sim \triangle \mathrm{ACB}$
(b) If $\mathrm{AC}=13 \mathrm{~cm}, \mathrm{BC}=5 \mathrm{~cm}$ and $\mathrm{AE}=4 \mathrm{~cm}$, find DE and AD .













