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°.