Question

please answer this......

Answer 1

HEYA!!!!

Here is your answer :
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ⅰ) CD is perpendicular to AB and CD = p

Area of triangle ABC = 1/2 × AB × CD

= 1/2cp.

also area of ∆ ABC = 1/2 × BC × AC

= 1/2AB

1/2cp = 1/2ab

let cp = ab --------------(1)

ii) Since ∆ABC is a right angled triangle at C.

${AB}^{2} = {BC}^{2} + {AC}^{2} \\ \\ {c}^{2} = {a}^{2} + {b}^{2} \\ \\ { \frac{ab}{p} }^{2} = {a}^{2} + {b}^{2} \\ \\ \frac{1}{ {p}^{2} } = \frac{ {a}^{2} + {b}^{2} }{ {a}^{2} {b}^{2} } \\ \\ \: \: \: \: \: \: \: = \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} }$

Hence proved...

HOPE THIS HELPS U...