Question

In the figure below, A, B, C, D are points on the circle.

Compute the angles of the quadrilateral ABCD and the angles between its diagonals.

Answer 1

We know that angles in a same segment are equal.

Here, ∠CBD and ∠CAD are the angles in the same segment.

⇒ ∠CBD = ∠CAD = 30°

Now, ∠BAD = ∠BAC + ∠CAD = 35° + 30° = 65°

∠ADB and ∠ACB are the angles in a same segment.

⇒ ∠ADB = ∠ACB = 50°

∠BDC and ∠BAC are the angles in a same segment.

⇒ ∠BDC = ∠BAC = 35°

Now, ∠ADC = ∠ADB + ∠BDC = 50° + 35° = 85°

We know that angles in the alternate segments are supplementary.

∴ ∠BAD + ∠BCD = 180°

⇒ 65° + ∠BCD = 180°

⇒ ∠BCD = 180° − 65° = 115°

Also, ∠ABC + ∠ADC = 180°

⇒ ∠ABC + 85° = 180°

⇒ ∠ABC = 180° − 85° = 95°

∴ ∠BAD = 65°, ∠ABC = 95°, ∠BCD = 115° and ∠ADC = 85°

Using angle sum property in ΔPBC:

∠PBC + ∠BCP + ∠CPB = 180°

⇒ 30° + 50° + ∠CPB = 180°

⇒ 80° + ∠CPB = 180°

⇒ ∠CPB = 180° − 80° = 100°

∠APD = ∠BPC = 100° (Vertically opposite angles)

We know that sum of angles forming a linear pair is 180°.

∴ ∠BPC + ∠BPA = 180°

⇒ 100° + ∠BPA = 180°

⇒ ∠BPA = 180° − 100° = 80°

∠CPD = ∠BPA = 80° (Vertically opposite angles)

Hence, the angles between the diagonals are ∠CPD = ∠BPA = 80° and ∠APD = ∠BPC = 100°.