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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.

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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

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

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  • $\begingroup$ You need to state $BD=PD$ or $OD$ is a common side for congruence, otherwise it is implicit assumption. $\endgroup$
    – farruhota
    Commented Sep 27, 2019 at 3:42
  • $\begingroup$ Yeah. Will add it. Thanks $\endgroup$
    – Quanto
    Commented Sep 27, 2019 at 3:49

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