Question

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 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

Answer 1

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 . <=== ! ! !