Question

In a given fig. $\angle ACB={90}^o$ and $CD\perp AB$ then which one of the following is true?

• A. $B{D}^2=AD\times CD$
• B. $A{D}^2=BD\times CD$
• C. $C{D}^2=BD\times AD$
• D. None of these

Answer 1

The correct option is C $C{D}^2=BD\times AD$

Let $\angle ACD=\theta$

$\Rightarrow \angle BCD=(90-\theta )$

which $\angle DBC=\theta$ $\{\angle CDB=90\}$

Now, In $△CDB,$ In $△ACD$

$\Rightarrow \tan \theta =\frac{CD}{BD}$ $\tan \theta =\frac{AD}{CD}$

equating both $\tan \theta$

$\Rightarrow \frac{CD}{BD}=\frac{AD}{CD}$

$\Rightarrow C{D}^2=BD\times AD$

Hence, the answer is $C{D}^2=BD\times AD.$