ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. Prove that ∠CBD + ∠CDB = 1/2 ∠BAD

Solution:

Given, ABCD is a quadrilateral.

A is the centre of the circle passing through B, C and D.

We have to prove that ∠CBD + ∠CDB = 1/2 ∠BAD

Join the diagonals AC and BD of the quadrilateral

Now, arc DC subtends an angle ∠DAC at the centre and ∠CBD at a point B in the remaining part of the circle.

We know that in a circle the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

So, ∠DAC = 2∠CBD ------------------ (1)

Similarly, arc BC subtends an angle ∠BAC at the centre and ∠CDB at a point D in the remaining part of the circle.

So, ∠BAC = 2∠CDB -------------------- (2)

Adding (1) and (2),

∠DAC + ∠BAC = 2∠CBD + 2∠CDB

From the figure,

∠DAC + ∠BAC = ∠DAB

So, ∠DAB = 2(∠CBD + ∠CDB)

Therefore, ∠CBD + ∠CDB = 1/2 ∠DAB

✦ Try This: ABCD is cyclic quadrilateral. The tangents to a circle at A and C meet at P. If ∠ APC = 50°, then the value of ∠ ADC is equal to ?

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10

NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 6

ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. Prove that ∠CBD + ∠CDB = 1/2 ∠BAD

Summary:

ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. It is proven that ∠CBD + ∠CDB = 1/2 ∠BAD

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