if a circle is drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
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Hello mate ☺
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Solution:
ABC is a triangle.
Taking side AC as diameter, we draw a circle and taking BC as diameter we draw another circle.
We need to prove that point D lies on the third side AB of ∆ABC.
AC is the diameter which means that ∠ADC=90° .,......(1)(Angle in a semi-circle is equal to 90°)
Similarly, BC is the diameter which means that ∠BDC=90° .........(2)(Angle in a semi-circle is equal to 90°)
From (1) and (2), we get ∠ADC+∠BDC=180° which means that ∠ADC and ∠BDC form a linear pair.
Therefore, points A, D and B are present on the same straight line.
I hope, this will help you.☺
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