Answer:
a. The required proof is obtained from triangles OAB and ABC formed by joining AB and that we have;
A[tex]\hat O[/tex]B + 2∠C = 180°, 2·A[tex]\hat P[/tex]B + 2∠C = 180°
∴ A[tex]\hat O[/tex]B = 2·A[tex]\hat P[/tex]B
b. i) P[tex]\hat S[/tex]S = P[tex]\hat R[/tex]Q, given that angles subtended by the same arc or chord are equal, therefore, in ΔPQS, we have;
[tex]\left | PS \right |[/tex] = [tex]\left |PQ \right |[/tex]
ii) S[tex]\hat Q[/tex]P = 45°, [tex]S\hat R Z[/tex] =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, A[tex]\hat O[/tex]B = 2·A[tex]\hat P[/tex]B
Let ∠O represent A[tex]\hat O[/tex]B and let ∠P' represent A[tex]\hat P[/tex]B
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;
A[tex]\hat O[/tex]B = 2·A[tex]\hat P[/tex]B
b. i) The given parameters are;
Points on the circle = P, Q, R, and S
P[tex]\hat Q[/tex]S = P[tex]\hat R[/tex]Q
According to circle theory, the angles which an arc or chord subtends in a given segment are equal, therefore;
P[tex]\hat S[/tex]S = P[tex]\hat R[/tex]Q
Therefore, P[tex]\hat S[/tex]S = P[tex]\hat Q[/tex]S by transitive property of equality
P[tex]\hat S[/tex]S and P[tex]\hat Q[/tex]S are base angles of ΔPQS, given that P[tex]\hat S[/tex]S = P[tex]\hat Q[/tex]S, we have;
ΔPQS is an isosceles triangle with base QS and therefore, the sides PS and PQ are the equal sides
Therefore, we have;
[tex]\left | PS \right |[/tex] = [tex]\left |PQ \right |[/tex]
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°
∴ [tex]S\hat PQ[/tex] = 90°
From (i), we have that P[tex]\hat S[/tex]S = P[tex]\hat Q[/tex]S, therefore, in triangle ΔPQS, we have;
[tex]S\hat PQ[/tex] + P[tex]\hat S[/tex]S + S[tex]\hat Q[/tex]P = 180°
Therefore;
90° + P[tex]\hat S[/tex]S + S[tex]\hat Q[/tex]P = 180°
P[tex]\hat S[/tex]S + S[tex]\hat Q[/tex]P = 180° - 90° = 90°
P[tex]\hat S[/tex]S = P[tex]\hat Q[/tex]S, therefore, P[tex]\hat S[/tex]S + S[tex]\hat Q[/tex]P = 2·S[tex]\hat Q[/tex]P
P[tex]\hat S[/tex]S + S[tex]\hat Q[/tex]P = 2·S[tex]\hat Q[/tex]P = 90°
S[tex]\hat Q[/tex]P = 90°/2 = 45°
S[tex]\hat Q[/tex]P = 45°
Similarly, given that SQ is the diameter, of the circle the angle [tex]S\hat R Q[/tex] formed by jointing S to Q is 90°
[tex]S\hat R Q[/tex] = 90°
[tex]S\hat R Q \ and \ S\hat R Z[/tex] are angles on a straight line and are therefore, supplementary, therefore;
[tex]S\hat R Z[/tex] = 180° - [tex]S\hat R Q[/tex]
[tex]S\hat R Z[/tex] = 180° - 90° = 90°
[tex]S\hat R Z[/tex] =90°.
