Construct triangle ABC, denote I as the incenter, A' as the mid point of the arc BC of the circumcircle. Show that A'B = A'C = A'I

Question

I know angles BAC and BCA are equal since the arcs BA and AC are equal however I do not know where to go on from there. Hints or answers involving cyclic quadrilaterals would be appreciated. Thank you.

Answer 1

Hint: since $A'$ is the midpoint of arc $BC$, it follows that $AA'$ is the bisector of $\angle BAC\,$, so the points $A, I, A'$ are collinear. Let $B'$ be the intersection of bisector $BI$ with the circumscribed circle.

Then:

• $\angle IBA' = \angle CBA'+ \angle IBC = \frac{\angle BAC}{2}+\frac{\angle ABC}{2}$ by the inscribed angle properties;

• $\angle BIA'= \frac{1}{2} \left(\text{arc}\, A'B + \text{arc}\, AB'\right) = \frac{\angle BAC+\angle ABC}{2}$ by the secant angle theorem.

Therefore $\angle IBA' = \angle BIA'\,$, so $\triangle IBA'$ is isosceles, so $A'B=A'I\,$. It follows in a symmetric way that $A'C=A'I\,$ which completes the proof.

Answer 2

The proof of the following fact is simple enough and hence is skipped.

In $\triangle ABC$, if I is the in-center, AI (the angle bisector of $\angle A$), when extended will meet the perpendicular bisector of $BC$ at A’. A’ will lie on the circum-circle of $\triangle ABC$. In short, AIA’ is a straight line and A’BAC is cyclic.

It should be clear that $\angle 1= \angle2$ and $\angle 3 =\angle 4$. Note also that $\angle 4 = \angle 5$.

$\angle A’IB =\angle 2 + \angle 3 = … = \angle A’BI$.

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Remark:- The circum-center is O, NOT I.