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CBSE Class 9 Mathematics Question 10 of 12

Circles — Question 10

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Question 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.

Answer

Two circles are drawn taking sides AB and AC as diameter.

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. NCERT Class 9 Mathematics CBSE Solutions.

⇒ ∠ADB = ∠ADC = 90° (Angles in the semi circle is a right angle.)

⇒ ∠ADB + ∠ADC = 90° + 90°

⇒ ∠ADB + ∠ADC = 180°

∴ BDC is straight line.

Thus, D lies on BC.

Hence, proved that the point of intersection of circles lie on the third side.

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Circles — Interactive Study Guide

Circle Theorems Quick Reference

  1. Equal chords ⇔ equal central angles
  2. Perpendicular from centre bisects the chord
  3. Equal chords ⇔ equidistant from centre
  4. Central angle = 2 × inscribed angle (same arc)
  5. Angles in same segment are equal
  6. Angle in semicircle = 90°
  7. Opposite angles of cyclic quad = 180°

Problem-Solving Toolkit

For chord problems: Drop perpendicular from centre → bisects chord → use Pythagoras.
For angle problems: Identify if angle is at centre or circumference → apply the 2x rule.
For cyclic quad: Opposite angles add up to 180°.

Quick Self-Check

  1. Inscribed angle = 35°. Central angle for the same arc? (70°)
  2. Chord = 10 cm, distance from centre = 12 cm. Radius? (√(25+144) = √169 = 13 cm)
  3. ABCD is cyclic, ∠A = 95°. Find ∠C. (85°)

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