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Question
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)
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