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

**Solution:**

We know that, the sum of all 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.

Consider ΔABC,

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

90° + ∠BCA + ∠CAB = 180°

∠BCA + ∠CAB = 90°.....(1)

Consider 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.

Since it is a cylclic quadrilateral the below figure can be drawn.

Consider chord CD. ∠CAD and ∠CBD are formed on the same segment CD.

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

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 10

**Video Solution:**

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

Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.5 Question 11

**Summary:**

ABC and ADC are two right triangles with common hypotenuse AC. We have found that ∠CAD = ∠CBD as they are lying on the same segment, and we know that they will always be equal

**☛ Related Questions:**

- If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
- If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
- Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D, P and Q respectively (see Fig. 10.40). Prove that ∠ACP = ∠QCD.
- If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

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