Given: A circle with center O. XY and X'Y' are two parallel tangents. AB is another tangent at point C, intersecting XY at A and X'Y' at B. P is the point of tangency for XY, and Q is the point of tan

Given: A circle with center O. XY and X'Y' are two parallel tangents. AB is another tangent at point C, intersecting XY at A and X'Y' at B. P is the point of tangency for XY, and Q is the point of tangency for X'Y'.
To prove: $\angle AOB = 90^\circ$.

We use the property that tangents drawn from an external point to a circle are equal in length, and the line segment joining the center to the external point bisects the angle between the tangents.

From external point A, AP and AC are tangents to the circle.
Therefore, $AP = AC$.
Consider $\triangle OPA$ and $\triangle OAC$:

• $OP = OC$ (radii of the same circle)
• $OA = OA$ (common side)
• $AP = AC$ (tangents from external point A) By SSS congruence criterion, $\triangle OPA \cong \triangle OAC$. Thus, by CPCTC (Corresponding Parts of Congruent Triangles are Congruent), $\angle POA = \angle COA$. Let $\angle COA = \alpha$. Then $\angle POA = \alpha$.

From external point B, BQ and BC are tangents to the circle.
Therefore, $BQ = BC$.
Consider $\triangle OQB$ and $\triangle OCB$:

• $OQ = OC$ (radii of the same circle)
• $OB = OB$ (common side)
• $BQ = BC$ (tangents from external point B) By SSS congruence criterion, $\triangle OQB \cong \triangle OCB$. Thus, by CPCTC, $\angle QOB = \angle COB$. Let $\angle COB = \beta$. Then $\angle QOB = \beta$.

Since XY and X'Y' are parallel tangents, the line segment PQ passing through the center O is a diameter. This means P, O, and Q are collinear.
Therefore, the angle $\angle POQ$ is a straight angle, i.e., $\angle POQ = 180^\circ$.
We can express $\angle POQ$ as the sum of the angles around O:

$$\angle POQ = \angle POA + \angle COA + \angle COB + \angle QOB$$

Substitute the values from Step 2 and Step 3:

$$180^\circ = \alpha + \alpha + \beta + \beta$$ $$180^\circ = 2\alpha + 2\beta$$ $$180^\circ = 2(\alpha + \beta)$$

Divide by 2:

$$\alpha + \beta = 90^\circ$$

The angle $\angle AOB$ is the sum of $\angle COA$ and $\angle COB$.

$$\angle AOB = \angle COA + \angle COB$$

Substitute the values from Step 2 and Step 3:

$$\angle AOB = \alpha + \beta$$

From Step 4, we found that $\alpha + \beta = 90^\circ$.
Therefore,

$$\angle AOB = 90^\circ$$