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)

Answer 2

Given: AC is the common hypotenuse, and \(\angle B = \angle D = 90^\circ\).
To prove: \(\angle CAD = \angle CBD\).

Proof:
1. Since \(\angle ABC\) and \(\angle ADC\) are both right angles (90°), and they are angles inscribed in a semicircle (half of a circle).
2. Any angle inscribed in a semicircle is always a right angle.
3. Thus, both triangles ABC and ADC lie on the circumference of a circle with AC as the diameter.
4. Therefore, points A, B, C, and D are concyclic (they lie on the same circle).
5. In a circle, if two chords (lines joining two points on the circumference) intersect, then the angles formed at the intersection are equal.
6. Hence, \(\angle CAD = \angle CBD\), as they are angles formed by intersecting chords AC and CD.

Therefore, we have proved that \(\angle CAD = \angle CBD\).