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

Δ ABC and Δ ADC are shown in the figure below:

From figure,

In ∆ ABC and ∆ ADC,

⇒ ∠B = 90° and ∠D = 90° [∵ ∆ ABC and ∆ ADC are right angled triangles]

We know that,

The sum of angles in a triangle is 180°.

If the sum of pair of opposite angles in a quadrilateral is 180°, then it is a cyclic quadrilateral.

In Δ ABC,

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

⇒ 90° + ∠BCA + ∠CAB = 180°

⇒ ∠BCA + ∠CAB = 180° - 90°

⇒ ∠BCA + ∠CAB = 90° …..(1)

In Δ ADC,

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

⇒ 90° + ∠ACD + ∠DAC = 180°

⇒ ∠ACD + ∠DAC = 180° - 90°

⇒ ∠ACD + ∠DAC = 90° …..(2)

Adding equation (1) and (2), we get :

⇒ ∠BCA + ∠CAB + ∠ACD + ∠DAC = 180°

⇒ (∠BCA + ∠ACD) + (∠CAB + ∠DAC) = 180°

⇒ ∠BCD + ∠DAB = 180° …..(3)

⇒ ∠B + ∠D = 90° + 90° = 180° …..(4)

Since, sum of opposite angles of quadrilateral ABCD is 180°. Therefore, it is a cyclic quadrilateral.

We know that,

Angles in the same segment are equal.

⇒ ∠CAD = ∠CBD.

Hence, proved that ∠CAD = ∠CBD.