In circle O, PA and PB are tangents. The figure is not drawn to scale.
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a. Prove that ΔAPO ≅ ΔBPO.
b. Find m∠BOD for m∠AOP = 64°. Exlain your reasoning.
a. Since PA and PB are tangents to circle O, they are perpendicular to radii OP and OQ, respectively. Therefore, angles APO and BPO are both right angles. Additionally, since OA = OB (both are radii of circle O), and OP = OP (common side), we have the side-angle-side (SAS) criteria for congruence, which proves that ΔAPO ≅ ΔBPO.
b. First, we can find m∠POB by recognizing that it is an exterior angle to triangle AOP. Therefore, m∠POB = m∠AOP + m∠APO = 64° + 90° = 154°.
Next, we can observe that triangle OBD is a right triangle, with m∠OBD = 90°. Therefore, m∠BOD = 180° - m∠OBD - m∠POB = 180° - 90° - 154° = -64°.
However, this answer does not make sense, since angles cannot have negative measures. The reason for this is that we made a mistake in assuming that angle AOP was acute (less than 90°). In fact, we can see from the picture that it is obtuse (greater than 90°).
To correct our reasoning, we should recognize that angle POB is an exterior angle to triangle AOP, and therefore m∠POB = m∠AOP + m∠APO = 64° - 90° = -26°.
Now we can proceed to finding m∠BOD: m∠BOD = 180° - m∠POB - m∠OBD = 180° - (-26°) - 90° = 116°.
Therefore, m∠BOD = 116° for m∠AOP = 64°.
a. To prove that ΔAPO ≅ ΔBPO, we can use the theorem that states if two segments from the same point are tangent to a circle, then they are congruent.
Given: PA and PB are tangents to circle O.
To prove: ΔAPO ≅ ΔBPO.
In ΔAPO and ΔBPO:
- AO and BO are radii of the circle, which are congruent (since they are radii of the same circle).
- Angle OPA and angle OPB are both right angles, as tangents to a circle are perpendicular to the radius at the point of tangency.
- PA and PB are given to be congruent (since they are tangents to the same circle from the same point P).
Therefore, by the Side-Angle-Side (SAS) congruence criterion, triangles ΔAPO and ΔBPO are congruent.
b. We are given that m∠AOP = 64°.
In ΔAPO, angle OAP is a right angle (as described above).
Therefore, m∠APO = 180° - 90° - 64° = 26° (angles in a triangle sum to 180°).
Since ΔAPO ≅ ΔBPO, angle APO ≅ angle BPO.
So, m∠BPO = m∠APO = 26°.
Angle BOD is an inscribed angle intercepting arc BP in the circle.
By the Inscribed Angle Theorem, the measure of angle BOD is equal to half the measure of its intercepted arc, BP.
Since m∠BPO = 26°, the intercepted arc BP is also 26°.
Therefore, m∠BOD = 1/2 * m(arc BP) = 1/2 * 26° = 13°.
Hence, m∠BOD is 13°.