ABC is a right traingle, right angled at C. if P is the length of perpendicular from C to AB and AB=c, BC=a and CA=b, then prove that (i) pc=ab (ii) `1/(p^(2)) = 1/(a^(2))+1/b^(2)`

Explanation

ABC is a right angled , at C, p is the lenth of the perpendicular from C to AB.
Proof: (i) `ar (triangleABC)=1/2xxAB xxCD`
`1/2cp`
`Also ar(triangleABC)=1/2 xxACxxBC`
= `1/2 ba …..(2) `
from (1) and (2) , we get
`1/2 pc= 1/2ab`
pc=ab ....(3)
(i) from (3) we get `c= (ab)/p`
`In triangleABC`
`c^(2)=a^(2)+b^(2) ` ( by Pythagoras theorem)
`((ab)/P)^(2) =a^(2)+b^(2) Rightarrow (a^(2)b^(2))/(P^(2)) = a^(2) + b^(2)`
`1/(P^(2))=(a^(2)+b^(2))/(a^(2)b^(2)) Rightarrow 1/(P^(2))= (a^(2))/(a^(2)b^(2))+b^(2)/(a^(2)b^(2))`
`1/P^(2)=1/(b^(2))+1/(a^(2))`
Alternative proof ( using trigonometry) let `angleCAB= theta`
`angleCBA= 90^(@)- theta (angleACB = 90^(@))`
In right `triangleCDA`
`sin theta= p/b`
in right `triangleCDB`
`sin (90-theta) = p/a`
`Rightarrow" " cos theta= p/a`
squaring and adding equations (1) and (2) , we get
` sin theta +cos^(2)theta= (p^(2))/(b^(2))+(p^(2))/(a^(2))`
`1= p^(2)((1/(a^(2))+1/(b^(2))) " " ( using identity sin^(2) theta+cos^(2) theta=1) `
`Rightarrow" " 1/(P^(2))= 1/(a^(2))+1/(b^(2))`
Remark : you will read `sin (90-theta) cos theta and sin^(2) theta + cos^(2) theta= 1 ` in the later chapter 8.