Question 32

ΔABC is right angled at C. 'p' is the perpendicular from C to AB; AB, BC, CA are c, a, b respectively, then:

Solution

$$\angle B\ is\ common\ and\ \angle BDC=\angle ACB=90^{\circ\ }$$

for $$\triangle ACB\ and\ \triangle BDC.$$

So, $$\triangle ACB\simeq\triangle BDC.$$

So, $$\frac{DC}{BC}=\frac{AC}{BA}.$$

or,$$\frac{p}{a}=\frac{b}{c}.$$

or,$$pc=ab.$$

C is correct choice.

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