Question

In figure, ABCD is a cyclic quadrilateral in which AB is extended to F and BE || DC. If

 =  FBE 20 and

 =  DAB 95 ,

then find

ADC.​

Answer 1

Given :

• ∠FBE = 20°
• ∠DAB = 95°
• ABCD is a cyclic quaderilateral.

To find :

• ∠ADC

According to the question,

⇒ ∠DAB + ∠BCD = 180°

Reason : Property of cyclic quaderilateral.

⇒ 95° + ∠BCD = 180°

⇒ ∠BCD = 180° - 95°

⇒ ∠BCD = 85°

⇒ BE || DC

⇒ ∠EBC = ∠BCD = 85°

⇒ ∠CBF = ∠EBF + ∠EBC

⇒ ∠CBF = 20° + 85°

⇒ ∠CBF = 105°

⇒ ∠CBF = ∠ADC

Reason : Angle in the same segment.

⇒ ∠ADC = 105°

So,

• ∠ADC = 105°

_______________________________

★ Some theorems :

Theorem 1 :

The angle subtended by an arc of a circle at the centre is doubled the angle subtended by it at any point on the remaining part of the circle.

Theorem 2 :

The angle in a semicircle is a right angle.

Theorem 3 :

Angles in the same segment of a circle are equal.

Answer 2

$\huge \bf Given$

ABCD is a cyclic Quadrilateral

$\sf \angle FBE = 20$

$\sf \angle DAB = 95$

$\huge \bf To \: Find$

ADC

$\huge \bf SoluTion$

Angle sum property of cyclic Quadrilateral is 180

$\sf \angle DAB + \angle BCD = 180$

$\sf \angle BCD = 180 - 95$

$\sf \angle BCD = 85$

$\sf \angle CBF = \angle ADC$

$\sf \angle CBF = 20 + 85$

$\frak \red{\angle CBF = 105 \degree}$

Therefore :-

$\sf \angle ADC = 105 \degree$