PUMPA - SMART LEARNING
எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
Book Free DemoA circle passing through the vertices
\(
A\) and
\(
B\) of a parallelogram
\(ABCD\) intersects the sides
\(AD\) and
\(BC\) at points \(X
\) and
\(Y\) respectively. Prove that the points \(X\),\(Y\),\(C\), and
\(D\) are concyclic.
Given: \(ABCD\) is a parallelogram.
Let a circle whose centre is \(O\) passes through \(A\) and \(B\) such that it intersects \(AD\) at \(X\) and \(BC\) at \(Y\).
Points \(X, Y, C\) and \(D\) are concyclic.
Now, Join point \(X\) to \(Y\).
Thus, \(XY\) line segment is constructed.
As, \(∠1=∠\) [Exterior angle property of cyclic quadrilateral]
But \(∠A=∠\) []
Therefore, \(∠1=∠C\) .....(1)
But \(∠C+∠D=\)\(^°\) []
\(∠1+∠D=\)\(^°\) [from (1)]
Therefore, the quadrilateral \(YCDX\) is cyclic.
So, the points \(X, Y, C\) and \(D\) are concyclic.
Hence, proved.
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