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.

Solution:

We know that an angle in a semicircle is a right angle. By using this fact, we can show that BDC is a line that will lead to proving that the point of intersection lies on the third side.

Since angle in a semicircle is a right angle, we get:

∠ADB = 90° and ∠ADC = 90°

∠ADB + ∠ADC = 90° + 90°

⇒ ∠ADB + ∠ADC = 180°

⇒ BDC is a straight line.

D lies on BC

Hence, the point of intersection of circles lies on the third side BC.

☛ Check: Class 9 Maths NCERT Solutions Chapter 10

Video Solution:

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

Maths NCERT Solutions Class 9 - Chapter 10 Exercise 10.5 Question 10

Summary:

If circles are drawn taking two sides of a triangle as diameters, we have found that the point of intersection of these circles lies on the third side.

☛ Related Questions:

• ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30° find ∠BCD. Further if AB = BC, find ∠ECD.
• If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
• If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
• Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D, P and Q respectively (see Fig. 10.40). Prove that ∠ACP = ∠QCD.