WXYZ is a cyclic quadrilateral.if bisectors of angle XWZ and angle angle XYZ meet the circle at the points P and Q respectively, then prove that PQ is the diameter of the circle.

To prove that PQ is the diameter of the circle in the context of a cyclic quadrilateral WXYZ, we need to rely on some properties of cyclic quadrilaterals and angle bisectors. Let's break this down step by step.

Understanding Cyclic Quadrilaterals

A cyclic quadrilateral is a four-sided figure where all vertices lie on the circumference of a circle. One of the fascinating properties of cyclic quadrilaterals is that the opposite angles are supplementary. This means that the sum of the measures of opposite angles is 180 degrees. For our quadrilateral WXYZ, we have:

• ∠WXY + ∠WZY = 180°
• ∠XYZ + ∠XWZ = 180°

Angle Bisectors and Their Intersection

Now, let’s focus on the angle bisectors of angles XWZ and XYZ. When we draw the angle bisector of an angle, it divides the angle into two equal parts. Thus:

• Let ∠XWZ be divided into two angles, ∠XWP and ∠PZW, where P is the point where the bisector meets the circumcircle again.
• Similarly, let ∠XYZ be divided into angles, ∠XYQ and ∠QZY, where Q is the point where the bisector meets the circumcircle again.

This means:

• ∠XWP = ∠PZW
• ∠XYQ = ∠QZY

Applying the Inscribed Angle Theorem

The Inscribed Angle Theorem states that an angle inscribed in a semicircle is a right angle. This is crucial in our proof. Since P and Q are points where the angle bisectors intersect the circumcircle, we can use this theorem effectively. Specifically, we want to show that the angle ∠PQW and ∠PQZ are right angles.

By the property of cyclic quadrilaterals, we know that:

• ∠WPQ + ∠WZQ = 180° (because they are opposite angles)
• ∠PQW = ∠WZY = ∠XYZ/2 (since P is on the bisector of XWZ)
• ∠PQZ = ∠XWY = ∠XWZ/2 (since Q is on the bisector of XYZ)

Establishing Right Angles

From the relationships we've established, we can summarize:

• ∠PQW = ∠XYZ/2
• ∠PQZ = ∠XWZ/2

Now, since ∠WXY + ∠WZY = 180°, we can substitute:

• ∠XWZ + ∠XYZ = 180°

This implies that:

• ∠PQW + ∠PQZ = 90° + 90° = 180°,

indicating that line segment PQ subtends a right angle at the center of the circle. Therefore, by the Inscribed Angle Theorem, PQ must be a diameter of the circle.

Final Thoughts

In conclusion, through the properties of cyclic quadrilaterals and the behavior of angle bisectors, we have established that the line segment PQ indeed acts as the diameter of the circumcircle of quadrilateral WXYZ. This elegant relationship highlights the beautiful connections within geometric figures!