Question

A B R and P are four points on a circle with centre 0. A O R and C are four points on a different circle. The two circles intersect at the points A and R . CPA CRB and AOB are straight lines. If angle CAB=x prove that angle ABC=x and reason on the Each statement must have a valid reason same line as the statement.​

Answer 1

Answer:

x° = ∠OBR = ∠ABC (base angles of a cyclic isosceles trapezoid)

Step-by-step explanation:

APRB form a cyclic trapezoid

∠APO  = x° (Base angle of an isosceles triangle)

∠OPR  = ∠ORP (Base angle of an isosceles triangle)

∠ORB = ∠OBR (Base angle of an isosceles triangle)

∠APO + ∠OPR + ∠OBR = 180° (Sum of opposite angles in a cyclic quadrilateral)

Similarly;

∠ORB + ∠ORP + x°  = 180°

Since ∠APO = x° ∠ORB = ∠OBR and ∠OPR  = ∠ORP we put

We also have;

∠OPR = ∠AOP = ∠BOR (Alternate interior angles of parallel lines)

Hence 2·x° + ∠AOP = 180° (Sum of angles in a triangle) = 2·∠OBR + ∠BOR

Therefore, 2·x° = 2·∠OBR, x° = ∠OBR = ∠ABC.