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ABCD is a cyclic quadrilateral in which AB = BC, O is the centre of circle and one of diagonal of cyclic quadrilateral passes through the centre of circle and ∠DCA = 40°. Find the ∠BAD.

ABCD is a cyclic quadrilateral in which AB = BC, O is the centre of circle and one of diagonal of cyclic quadrilateral passes through the centre of circle and ∠DCA = 40°. Find the ∠BAD. Correct Answer 95° 

Given:

ABCD is a cyclic quadrilateral

AB = BC

'O' is the center of a circle

∠DCA = 40° 

Concept used:

In the cyclic quadrilateral, the sum of the opposite angle is 180° 

And angle made in a semi-circle through its diameter is 90° 

The sum of all angle inside a triangle = 180° 

In a triangle, if two sides are equal, it will be an isosceles triangle and also two angle opposite to the two equal sides are equal.

Calculation:

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Let us assume AC be diagonal that passes through the center

From the above figure, we can see that AC is diagonal, and AC is also a diameter of circle then ∠ABC = ∠ADC = 90°  

In ΔADC,

∠CAD + ∠ADC + ∠DCA = 180°

⇒ ∠CAD + 90° + 40° = 180°

⇒ ∠CAD + 130° = 180° 

⇒ ∠CAD = 180° - 130° 

⇒ ∠CAD = 50° 

In ΔABC, it is isosceles triangle because AB is equal to BC, then ∠BAC = ∠ACB 

∠BAC + ∠ABC + ∠BCA = 180° 

⇒ 2∠BAC + 90° = 180° 

⇒ 2∠BAC = 180° - 90° 

⇒ 2∠BAC = 90° 

⇒ ∠BAC = 90°/2

⇒ ∠BAC = 45°

∠BAD = ∠BAC + ∠CAD

⇒ ∠BAD = 45° + 50° 

⇒ ∠BAD = 95° 

∴ The ∠BAD is 95°               

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