Q: 10    If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Answers (1)

Given: circles are drawn taking two sides of a triangle as diameters.

Construction: Join AD.

Proof: AB is the diameter of the circle and \angleADB is formed in a semi-circle.

             \angleADB = 90 \degree........................1(angle in a semi-circle)

Similarly,          

AC is the diameter of the circle and \angleADC is formed in a semi circle.

             \angleADC = 90 \degree........................2(angle in a semi-circle)

From 1 and 2, we have

      \angleADB+\angleADC=90 \degree+90 \degree=180 \degree

\angleADB and \angleADC are forming a linear pair. So, BDC is a straight line.

Hence, point D lies on this side.

 

 

 

 

 

  

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