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a. The required proof is obtained from triangles OAB and ABC formed by joining AB and that we have;

AB + 2∠C = 180°, 2·AB + 2∠C = 180°

∴ AB = 2·AB

b. i) PS = PQ, given that angles subtended by the same arc or chord are equal, therefore, in ΔPQS, we have;

=

ii) SP = 45°, =90°

Step-by-step explanation:

a. The given parameters are;

The center of the circle is point O

Points on the circumference of the circle = A, B, and P

Required to be proved, AB = 2·AB

Let ∠O represent AB and let ∠P' represent AB

We draw a line from the center O to the point P, and a line joining points A and B on the circumference of the circle

In ΔOAB, we have;

∠O + 2∠C = 180° (The sum of the interior angles of a triangle)

In ΔAPB, we have;

∠P' + ∠(C - a) + ∠(P' + C + a) = 180°

∴ 2·∠P' + 2·∠C = 180°

Therefore, by addition property of equality, we get;

∠O = 2·∠P'

Therefore;

AB = 2·AB

b. i) The given parameters are;

Points on the circle = P, Q, R, and S

PS = PQ

According to circle theory, the angles which an arc or chord subtends in a given segment are equal, therefore;

PS = PQ

Therefore, PS = PS by transitive property of equality

PS and PS are base angles of ΔPQS, given that PS = PS, we have;

ΔPQS is an isosceles triangle with base QS and therefore, the sides PS and PQ are the equal sides

Therefore, we have;

=

ii) Given that SQ is the diameter of the circle, we have by circle theorem, the angle subtended on the circumference by the diameter = 90°

∴ = 90°

From (i), we have that PS = PS, therefore, in triangle ΔPQS, we have;

+ PS + SP = 180°

Therefore;

90° + PS + SP = 180°

PS + SP = 180° - 90° = 90°

PS = PS, therefore, PS + SP = 2·SP

PS + SP = 2·SP = 90°

SP = 90°/2 = 45°

SP = 45°

Similarly, given that SQ is the diameter, of the circle the angle formed by jointing S to Q is 90°

= 90°

are angles on a straight line and are therefore, supplementary, therefore;

= 180° -

= 180° - 90° = 90°

=90°.

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step-by-step explanation:

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