Question

In the given figure $\angle ABC={90}^0$ and BD ⊥ AC. If $BD=8cm$ and $AD=4cm$, find $CD$.

Answer 1

Given: \( \begin{array}{l} \angle BDC =90^{\circ} \\ \angle ABC =90^{\circ} \end{array} \) Let \( \angle BCD = x ^{\circ} \) Using angle sum property of a triangle in \( \triangle BDC , \angle DBC =90^{\circ}- x ^{\circ} \) Also \( \angle ABD = x ^{\circ}\left(\angle D B C=90^{\circ}-x^{\circ}\right) \) Using angle sum property of a triangle in \( \triangle ADB , \angle BAD =90^{\circ}- x ^{\circ} \) Now considering \( \triangle BDC \) and \( \triangle ADB \) \( \begin{array}{l} \angle BDC =\angle BDA \\ {\left[\because \angle BDC =\angle BDA =90^{\circ}\right]} \\ \angle DBC =\angle DAB \\ {\left[\because \angle DBC =\angle DAB =(90-x)^{\circ}\right]} \end{array} \) So by AA \( \triangle BDC \sim \triangle ADB \)
Hence \( \frac{ BD }{ CD }=\frac{ AD }{ BD } \) \( \frac{8}{C D}=\frac{4}{8} \) \( C D=16 \) Hence \( C D=16 cm \)