Abc is a right angled triangle at c

Asked by admin @ 31/08/2021 in Math viewed by 357 People

abc is a right angled at c.if p is the length of the perpendicular from c to ab and a,b,c have the usual meaning then prove that 1/p^2=1/a^2+1/b^2

Answered by admin @ 31/08/2021

Answer:

Here, ABC is a right triangle,

In which ∠C = 90°, AB =c, AC= b, BC = a,

Also, p is the length of the perpendicular from c to ab,

We have to prove that:

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

Proof:

Let D be the point on the segment AB,

Such that CD = p,

In triangle ACB and ADC,

$\angle CAB\cong \angle DAC$ ( reflexive )

$\angle ACB\cong \angle ADC$ ( Right angle )

By AA similarity postulate,

$\triangle ACB\sim \triangle ADC$

By the property of similar triangles,

$\frac{AC}{CB}=\frac{AD}{DC}$

$\frac{b}{a}=\frac{AD}{p}$

$\frac{pb}{a}=AD$ ------------ (1)

Now, by the Pythagoras theorem,

$AC^2=AD^2+CD^2$

$b^2=(\frac{pb}{a})^2+p^2$

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

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

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

Hence, proved.

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