Question

In the given figure ∠BAC = 90° and AD ⊥ BC. Then,


(a) BC ⋅ CD = BC²
(b) AB ⋅ AC = BC²
(c) BD ⋅ CD = AD²
(d) AB ⋅ AC = AD²

Answer 1

(c) BD ⋅ CD = AD²

$$\begin{gathered} \mathrm{In}△BDA\mathrm{and}△ADC,\mathrm{we}\mathrm{have}: \\ \angle BDA=\angle ADC=90° \\ \angle ABD=90°-\angle DAB \\ =90°-\left(90°-\angle DAC\right) \\ =90°-90°+\angle DAC \\ =\angle DAC \\ \mathrm{Applying}AA\hspace{0.17em}\mathrm{similarity}\mathrm{theorem},\mathrm{we}\mathrm{conclude}\mathrm{that}△BDA~△ADC. \\ \Rightarrow \frac{BD}{AD}=\frac{AD}{CD} \\ \Rightarrow A{D}^2=BD.CD \end{gathered}$$