In the picture, O is the centre of the circle and A, B, C are points on it. Prove that ∠ OAC + ∠ ABC = 90°.

Case1: The angle got by joining any point on the larger part of the circle to the end of the chord is half the angle got by joining the centre of a circle to these end.

● Join A to the centre O of the circle. This line splits the angle at A into two parts, x° and y° .

● Here Δ AOB and Δ AOC are isosceles triangle

Thus, ∠ ABO = ∠ OAB = x° and ∠ ACO = ∠ OAC = y°

● Let us take ∠ BOC = c°

⇒ (180° – 2x) + (180° – 2y) + c° = 360°

⇒ 360° –2 (x+y) +c° = 360°

⇒ 2(x+y) = c°

Thus,

Consider, the diagram given in question.

Join, OA and OB

In, Δ OAB

Since, OA and OB are radius of a circle

Therefore, ∠ OAB = ∠ OBA (since, angles opposite to

equal side are equal) ..(1)

By, angle sum property,

∠ OAB + ∠ OBA + ∠ AOB = 180°

⇒ ∠ OAB + ∠ OAB + ∠ AOB = 180° (from (1))

⇒ 2 ∠ OAB = 180° – ∠ AOB

….(2)

As, Δ OAC and Δ OBC are isosceles

∴ ∠ ACO = ∠ CAO and ∠ OBC = ∠ BCO …..(3)

Also, by case1 we get,

…(4)

Consider,

∠ CAO + ∠ ABC = ∠ CAO + ∠ CBO + ∠ OBA

(from (2) and (3))

(from (4))

= 90°