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Since $OB = OP$, $\angle OBD = \angle OPD = 90^\circ$, and the common side $OD$, the triangles $\triangle OBD$ and $\triangle OPD$ are congruent. Thus,

$$\angle BOD = \angle POD$$

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

$$\angle POE = \angle COE$$

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

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

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

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

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

$$\angle BOD = \angle POD$$

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

$$\angle POE = \angle COE$$

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

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

Since $OB = OP$, $\angle OBD = \angle OPD = 90^\circ$, and the common side $OD$, the triangles $\triangle OBD$ and $\triangle OPD$ are congruent. Thus,

$$\angle BOD = \angle POD$$

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

$$\angle POE = \angle COE$$

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

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

Since $OB = OP$ and $\angle OBD = \angle OPD = 90^\circ$, the triangle $\triangle OBD$ and $\triangle OPD$ are congruent. Thus,

$$\angle BOD = \angle POD$$

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

$$\angle POE = \angle COE$$

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

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

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

$$\angle BOD = \angle POD$$

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

$$\angle POE = \angle COE$$

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

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