Tangents at B, C and P on a circle O intersect at points D and E. Prove $\angle DOE = \frac 12 \angle BOC$

Question

The tangents to the circle at B and C meet at the point A. A point P is located on the minor arc BC and the tangent to the circle at P meets the lines AB and AC at the points D and E, respectively. Prove that

$$\angle DOE = \frac 12 \angle BOC$$

where O is the center of the given circle.

How should I start this question? What identities or theorems should I use?

Thanks in advance for helping me out.

Answer 1

Since $OB = OP$, $\angle OBD = \angle OPD = 90^\circ$, and the common side $OD$, $\triangle OBD$ and $\triangle OPD$ are congruent, which leads to

$$\angle BOD = \angle POD$$

Similarly, $\triangle POE$ and $\triangle COE$ are congruent, leading to

$$\angle POE = \angle COE$$

Therefore, $\angle DOP = \frac 12 \angle BOP$ and $\angle EOP = \frac 12 \angle COP$, yielding

$$\angle DOE = \frac 12 \angle BOC$$