# Another Property of Circle Through the Incenter

### Problem

$I$ is the incenter of $\Delta ABC.$ Let $BE$ be a chord in $(ABI)$ such that $BE=AB.$ Let $O$ be the circumcenter of $\Delta ABC.$

Then $BE\perp BO.$

### Proof

The proof follows by angle chasing. (The angles are as shown in the diagram.)

Let $O_c$ be t he center of $(ABI),$ $BG$ its diameter, and $BH$ a diameter of $(ABC).$

We know that

- $\angle GBA=\angle O_{c}BA=\gamma/2.$
- $BE$ is a reflection of $AB$ in the line $BO_c,$ i.e., $BG,$ implying that $\angle ABE=\gamma.$
- $\angle AHB=\angle ACB=\gamma$ so that $\angle ABO=90^{\circ}-\gamma.$

It follows that $\angle EBO=\angle EBA+\angle ABO=\gamma+(90^{\circ}-\gamma)=90^{\circ}.$

### Acknowledgment

The problem stemmed from the one posted by Dao Thanh Oai at the at the CutTheKnotMath facebook page.

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