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First Four Postulates

Q.ln triangle...

Question

Q.ln triangle ABC,right angled at C, if p is the length of the perpendicular from C to AB and a, b ,c have their usual meanings, then prove that A. cp=ab B. 1/p2 = 1/a2 + 1/b2

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\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

c p = a b

\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

A B C

C

B C = a, C A = b, A B = c

p

C

A B

c p = a b

\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

\frac{1}{a^{2}} + \frac{1}{b^{2}}

A B C

C .

B C = a, C A = b, A B = c

p

C

A B

p c = a b

\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

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