# PLEASE HELP ME WITH IT. In the figure given below , AB and AC are two chords of a circle of center O and radius r. If AB= 2 AC and the perpendiculars drawn fro

Mathematics

## Question

In the figure given below , AB and AC are two chords of a circle of center O and radius r. If AB= 2 AC and the perpendiculars drawn from the center on these chords are of lengths 'a' and 'b' respectively. PROVE THAT 4b^2= a^2+3r^2

This question is related to lesson CIRCLES

• ### 1. User Answers AL2006

OK.  I did it.  Now let's see if I can go through it without
getting too complicated.

I think the key to the whole thing is this fact:

A radius drawn perpendicular to a chord bisects the chord.

That tells us several things:

-- OM bisects AB.
'M' is the midpoint of AB.
AM is half of AB.

-- ON bisects AC.
'N' is the midpoint of AC.
AN is half of AC.

--  Since AC is half of AB,
AN is half of AM.
a = b/2

Now look at the right triangle inside the rectangle.
'r' is the hypotenuse, so

a² + b² = r²

But  a = b/2, so             (b/2)² + b² = r²

(b/2)² = b²/4                   b²/4   + b² = r²

Multiply each side by 4:     b² + 4b² = 4r²
-  -  -  -  -  -  -  -  -  -  -
0  + 5b² = 4r²
Repeat the
original equation:                a² +  b² =  r²

Subtract the last
two equations:                  -a² + 4b² = 3r²

Add  a²  to each side:              4b²  =  a² + 3r² .    <=== ! ! !