In triangle ABC, ∠C=90°. Let BC= a, CA= b, AB= c and let 'p' be the length of the perpendicular from 'C' on AB, prove that:

1. cp = ab

2. `1/p^2=1/a^2+1/b^2`

#### Solution

1. Area of a triangle = (1/2) x Base x Height

A(ΔABC) = (1/2) x AB x CD

A(ΔABC) = (1/2) x cp .......(i)

Area of right angle triangle ABC = A(ΔABC) = (1/2) x AC x BC

A(ΔABC) = (1/2) x ba ........(ii)

From (i) and (ii)

cp=ba⇒ cp⇒ab ..........(iii)

2. We have,

cp=ab ..........From(iii)

p = ab/c

Square both sides of the equation.

We get, `p^2=(a^2b^2)/c^2`

`1/p^2=c^2/(a^2b^2)" ..................(iv).....[By invertendo]"`

In right angled triangle ABC,

AB^{2} = AC^{2} + BC^{2} ................[By Pythagoras’ theorem]

c^{2} = b^{2} + a^{2} ............(v)

`c^2/(a^2b^2) = b^2/(a^2b^2) + a^2/(a^2b^2)`..........[Dividing throughout by `a^2b^2`]

`c^2/(ab)^2 = 1/a^2 + 1/b^2` .........(iii)

`c^2/(cp)^2 = 1/a^2 + 1/b^2` ...........[From (ii) and (iii)]

`c^2/(c^2p^2) = 1/a^2 + 1/b^2`

`1/p^2 = 1/a^2 + 1/b^2`