Question

In the given figure , the common tangents AB and CD to two circles with centres O and O' intersect at E . Prove that the point O , E and O' are collinear .

Answer 1

Construction: Join OA & OC.

In $△OAE$ & $△OCE$, we have

$OA=OC$ [radii of the same circle]

$OE=OE$ [common side]

$\angle OAE=\angle OCE$

$△OAE\cong △OCE$ [RHS congruence criterion]

$\angle AEO=\angle CEO$ [By CPCT]

Similarly,

$\angle BE{O}^{\prime }=\angle DE{O}^{\prime }$

$\angle AEC=\angle DEB$

$\frac{1}{2}\angle AEC=\frac{1}{2}\angle DEB$

$⟹\angle AEO=\angle CEO=\angle BE{O}^{\prime }=\angle DE{O}^{\prime }$

Since, these angles are equal and are bisected by OE & O'E,

$O,E,O^{\prime }$ are collinear.