Wxyz is a cyclic quard. If the bisector of angle xwz and angle xyz meet the circle at point a and b resp. then prove:ab is a diameter of a circle
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Answer
To prove that AB is a diameter of the circle, we need to show that angle AOB is a right angle, where O is the center of the circle.
Since WXYZ is a cyclic quadrilateral, this means that opposite angles are supplementary. Therefore, angle XYZ + angle WXY = 180 degrees.
Let M be the midpoint of the minor arc XYZ (arc YZ). Since the bisector of angle XYZ meets the circle at point A, this means that angle XAM = angle MAY. Similarly, since the bisector of angle XWZ meets the circle at point B, angle WBZ = angle ZBM.
Now, let's consider triangle XZW. Since angle XZW is a straight angle (180 degrees), we have:
angle XZW + angle XYZ + angle WXY = 180 degrees
angle XZW + (angle XAM + angle MAY) + (angle WBZ + angle ZBM) = 180 degrees
Simplifying this expression, we get:
angle XZW + angle XAM + angle MAY + angle WBZ + angle ZBM = 180 degrees
Since angle XZW and angle XAM are supplementary (because they share the same bisector), angle XZW + angle XAM = 180 degrees. Similarly, angle WBZ + angle ZBM = 180 degrees.
Therefore, we can rewrite the expression as:
180 degrees + angle MAY + 180 degrees + angle MAY = 180 degrees
Simplifying further:
360 degrees + 2 * angle MAY = 180 degrees
Subtracting 360 degrees from both sides:
2 * angle MAY = -180 degrees
Dividing both sides by 2:
angle MAY = -90 degrees
Since angle MAY is a negative angle, we can consider its supplement, which is angle AMO (where O is the center of the circle). Therefore, angle AMO = 90 degrees.
This means that angle AOB is also a right angle, since it is the sum of angles AMO and BMO. Therefore, AB is a diameter of the circle QED (which stands for "quod erat demonstrandum" and means "which had to be proven").
To prove that AB is a diameter of the circle, we need to show that angle AOB is a right angle, where O is the center of the circle.
Using the properties of a cyclic quadrilateral, we know that the opposite angles are supplementary. Therefore, angles XWZ and XYZ are supplementary.
Let's consider angle XWZ. Since the bisector of angle XWZ intersects the circle at point A, we have angle XWA = angle ZWA.
Now let's consider angle XYZ. Since the bisector of angle XYZ intersects the circle at point B, we have angle XYB = angle ZYB.
Since angles XWZ and XYZ are supplementary, we can write the equation: (angle XWA + angle ZWA) + (angle XYB + angle ZYB) = 180°.
Rearranging the equation, we have: (angle XWA + angle XYB) + (angle ZWA + angle ZYB) = 180°.
But angle XWA is equal to angle XYB and angle ZWA is equal to angle ZYB.
Therefore, the equation becomes: 2(angle XWA) + 2(angle ZWA) = 180°.
Dividing both sides of the equation by 2, we get: (angle XWA + angle ZWA) = 90°.
This means that angle AOB is a right angle, proving that AB is a diameter of the circle.