Consider the diagram shown below.
Given:
AB and AC are two chords of a circle with center O. Such that AB=2AC
p and q are ⊥ distances of AB and AC from center O i.e., OM=p and ON=q
r is the radius of the circle
To prove that:
4q2=p2+3r2
Proof:
Join OA.
OM and ON are ⊥ distances of AB and AC from center O.
Here,
AN=AC2 (perpendicular from center to chord intersect at mid-point of the chord)
AM=AB2 (perpendicular from center to chord intersect at mid-point of the chord)
In right angled ΔOMA,
OM2+AM2=OA2
p2+AM2=r2
AM2=r2−p2 …… (1)
In right angled ΔONA,
ON2+AN2=OA2
q2+AN2=r2
AN2=r2−q2 …… (2)
Since,
AM=AB2=2AC2=AC=2AN
From equations (1) and (2), we have
r2−p2=AM2
r2−p2=4AN2
r2−p2=4[r2−q2]
4q2=p2+3r2
LHS=RHS
Hence, proved.