Question

In Fig. 7.238, ∠BAC = 90° and AD ⊥ BC. Then, BD. CD = _________.

Answer 1

In ∆ABD,

∠ABD + ∠BAD = 90º .....(1)

Now, ∠BAC = 90°

Or ∠CAD + ∠BAD = 90º .....(2)

From (1) and (2), we have

∠ABD + ∠BAD = ∠CAD + ∠BAD

⇒ ∠ABD = ∠CAD

In ∆ABD and ∆ACD,

∠ADB = ∠ADC (90º each)

∠ABD = ∠CAD (Proved above)

∴ ∆ABD $~$ ∆CAD (AA Similarity)

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{CA}}=\frac{\mathrm{AD}}{\mathrm{CD}}=\frac{\mathrm{BD}}{\mathrm{AD}}$ (If two triangles are similar, then their corresponding sides are proportional)

$$\begin{gathered} \Rightarrow \frac{\mathrm{AD}}{\mathrm{CD}}=\frac{\mathrm{BD}}{\mathrm{AD}} \\ \Rightarrow \mathrm{BD}.\mathrm{CD}={\mathrm{AD}}^2 \end{gathered}$$

In Fig. 7.238, ∠BAC = 90° and AD ⊥ BC. Then, BD. CD = ____AD²____.