6. Exemplar - Prove the following I

Explanation:

 

Now, Join \(YD\) and \(YC\).

 

 

Here, \(ABCD\) is a cyclic quadrilateral.

 

The bisectors of opposite angles of the cyclic quadrilateral, \(∠A\) and \(∠C\), intersect the circle circumscribing at the points \(X\) and \(Y\) respectively.

 

Now, the opposite angles of a cyclic quadrilateral are supplementary.

 

That is, \(∠B + ∠D = \)\(^°\)

 

\(\frac{1}{2}∠B+\frac{1}{2}∠D=\frac{1}{2} \times \)\(^°\)

 

\(=\)\(^°\) 

 

That is,\(\angle XDY=\)\(∠CDX + \)\(CBY\)\(XDY\)\(CDY\)\(= 90^°\)

 

But \(∠CBY =\)\(CDY\)\(AXY\)\(XDY\) [Angles subtends by a diameter of the circle are equalAngles in the same segment of a circle are equal subtended by the chord \(QD\)Angles in the same segment of a circle are equal subtended by the chord \(QC\)] 

 

Therefore, \(∠CDX + ∠CDY = \)\(^°\) 

 

\(∠\)\(XDY\)\(AXY\)\(CDY\)\(= 90^°\)

 

Thus, \(∠XDY\) is in semicircle.  [The diameter of the circle subtends a right angle at the circumference]

 

Therefore, \(XY\) is diameter of the circle.

 

Hence, proved.