Question

A chord AB of a circle, centre O, is produced to C. The straight line bisecting $\angle OAB$ meets the circle at E. Prove that EB bisects $\angle OBC$

Answer 1

Given: $\angle OAE=\angle EAB$

$\angle OAB=\frac{1}{2}\angle OAE=\frac{1}{2}\angle EAB$

$\angle EBC$ is an external angle to $△AEB$

$\angle EAB+\angle AEB+\angle EBC–\left(1\right)$

WKT angle made by a vhord at center is twice the angle made by the same chord on the circumference

$⟹\angle AOB=2\angle AEB$

$\angle OBC$ is an external angle to $△OAB$

$⟹\angle OAB+\angle APB=\angle OBC$

$\frac{1}{2}\angle EAB+\frac{1}{2}\angle AEB=\angle OBC$

From $\left(1\right)$

$\frac{1}{2}\angle EBC=\angle OBC⟹EBbisects\angle OBC$