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High School Mathematics - 2
8.4 Theorems - Circles

Theorem: Angle is a semi circle is a right angle.

Given : Angle AOB is in semi circle.
To prove: Angle ACB = 90o
Proof: Angle ACB = 1/2 angle AOB (angle subtended at the centre is twice angle at circumference)
Angle AOB = 180o (straight line) Hence ACB = 90o

Theorem: Angles in the same segment of a circle are equal.


Directions: Solve the following

Name: ___________________

Date:___________________

High School Mathematics - 2
8.4 Theorems - Circles

Q 1: Find x.

140 degrees
70 degrees
35 degrees

Q 2: If diagonals AC and BD of a quadrilateral intersect at M. Prove that a line drawn through M to bisect any side of the quadrilateral is perpendicular to the opposite side.
Answer:

Q 3: ABC is a triangle and P is a point on BC such that AB = BP. If AQ is produced to meet the circumcircle of triangle ABC at Q. Prove that CP = CQ
Answer:

Q 4: find x.

55 degrees
110 degrees
90 degrees

Q 5: Two circles intersect each other at points A and B. If AP and BQ be the respective diameters, show that PBQ is a straight line.
Answer:

Q 6: In the figure, CAD and CBE are straight lines. If CA is the diameter of the circle ABC, find angle ADE

90 degrees
180 degrees
45 degrees

Q 7: In the figure, PAQ and RAS are straight lines. Show that angPXR = angQYS.

Answer:

Q 8: ABCD is a cyclic quadrilateral and BC = CD. Show that AC bisects –BAD

Answer:

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Question 10: This question is available to subscribers only!


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