Question

ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠ CBD.

Answer 1

In ∆ABC,

∠ABC + ∠BCA + ∠CAB = 180°

90° + ∠BCA + ∠CAB = 180°

∠BCA + ∠CAB = 90°.....(1)

∠ABC +∠ BCA + ∠CAB = 180° (Angle sum property of a triangle)

In ∆ADC,

∠CDA + ∠ACD + ∠DAC = 180° (Angle sum property of a triangle)

90° + ACD + DAC = 180°

∠ACD + ∠DAC = 90°.....(2)

Adding equations (1) and (2), we obtain

∠BCA + ∠CAB + ∠ACD + ∠DAC = 180°

(∠BCA +∠ACD) + (∠CAB + ∠DAC) = 180°

∠BCD + ∠DAB = 180° ... (3)

However, it is given that

∠B + ∠D = 90° + 90° = 180°.....(4)

From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180°. Therefore, it is a cyclic quadrilateral.

Consider chord CD.

∠CAD = ∠CBD (Angles in the same segment)