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

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#### Solution

In ΔABC,

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

⇒ 90° + ∠BCA + ∠CAB = 180°

⇒ ∠BCA + ∠CAB = 90° ... (1)

In ΔADC,

∠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)

Concept: Cyclic Quadrilateral

Is there an error in this question or solution?

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