Question

Two circle with centres O and O' are drawn to intersect each other at points A and B.

Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O' at A. Prove that OA bisects $\angle BAC$

Answer 1

GIVEN: Circles with centre O & O' , AC is tangent to the circle with centre o' at A. OO' meets AB at E.

TO PROVE THAT: $\angle BAO=\angle CAO$

PROOF: In $△O^{\prime }AO$

O'A = O'O ( being radii of the same circle with centre O')

So, $\angle O^{\prime }AO=\angle O^{\prime }OA$ = y……..(1)

O'A is perpendicular to tangent AC ( as theorem states that tangent to any circle at any point is perpendicular to the radius through that point.)

So, $\angle O^{\prime }AC$= 90°…………..(2)

Eq(2) - Eq(1)

$\angle O^{\prime }AC—\angle O^{\prime }AO$

= $\angle OAC$ = 90°— y ………..(3)

Now since ,

O is equidistant from A& B ( being radii of the same circle with centre O)

That implies that O lies on the perpendicular bisector of the segment AB.

Similarly, O' is equidistant from A & B( being radii of the same circle with centre O')

That implies that O' lies on the perpendicular bisector of the segment AB.

Hence conclude that OO' is perpendicular bisector of AB , at E ( by theorem )

Now, in $△AEO$

$\angle AEO$ = 90° ( proved above)

$\angle O^{\prime }OA=y$ ( by eq (1) )

Therefore, third $\angle OAE=90°—y\dots \dots \dots \left(4\right)$

By comparing eq(3) & (4)

We have, $\angle OAC=\angle OAE$

[Hence proved]