NABTEB Mathematics 2024 Theory — Question 4

Model answer

Since OBCD is a rhombus and O is the centre of the circle passing through A, B, C, D, all sides OB, BC, CD, DO equal the radius r, and since C also lies on the circle, OC = r too. This means triangles OBC and OCD are each equilateral (all three sides = r), so ∠BOC = ∠COD = 60°, giving ∠BOD (measured through C) = 120°. (a) The angle t = ∠BAD is the inscribed angle standing on arc BD (the arc containing C), and the angle at the centre standing on the same arc is 120°. By the inscribed angle theorem (angle at centre = twice angle at circumference): t = 120°/2 = 60°. (b) Since OA=OB=r, triangle OAB is isosceles with ∠OAB=∠OBA=y. Similarly triangle OAD is isosceles with ∠OAD=∠ODA=y. Since OA bisects ∠BAD by symmetry: t = 2y → 60° = 2y → y = 30°. (c) In the rhombus OBCD, opposite angles are equal: ∠O=∠C=120°, so ∠B+∠D=360°−120°−120°=120°, and since ∠B=∠D, each = 60°. So ∠ODC=60°. ∠ADC = ∠ADO + ∠ODC = y + 60° = 30° + 60° = 90°.