Question

In figure, AB and DC of cyclic quadrilateral ABCD are produced to meet at E sides AD and BC are produced to meet at F If angle ADC = 80° and angle BAD = 50° then find angle BAD and angle CFD.​

Answer 1

1.ADE+EAD+AED=180°

80°+EAD+ 50°=180

EAD=50°=BAD

2.D+B=180°

80+B=180°

ABC=100°

NOW, IN ∆ABF

A+B+F=180°

50°+100°+F=180°

F=30°=CFD

Answer 2

Answer:

$\tt \:  \longrightarrow \: \angle \: BAD = 50 \degree \: \\ \tt \:  \longrightarrow \: \angle \: CFD = 30 \degree \: \:$

Step-by-step explanation:

Properties used :-

• 1. Ange sum property of triangle states that the sum of interior angles of a triangle is 180°.

• 2. The sum of the opposite angles of a cyclic quadrilateral is upplementary. 

Solution:-

☆ In triangle ️ AED

☆ Using angle sum property of triangle,

$\tt \:  \longrightarrow \: \angle \: AED + \angle \: EDA + \angle \:EAD = 180$

$\tt \:  \longrightarrow \: \angle \: EAD + 50 + 80 = 180$

$\tt \:  \longrightarrow \: \angle \: EAD + 130 = 180$

$\tt \:  \longrightarrow \: \angle \: EAD = 180 - 130$

$\tt \:  \longrightarrow \: \angle \: EAD = 50$

$\bf\implies \: \boxed{\tt \:  \green{\angle \:BAD \: = \: 50}}$

☆ In cyclic quadrilateral ABCD,

☆ Using sum of the opposite angles of cyclic quadrilateral is 180°.

$\tt \:  \longrightarrow \: \angle \:ADC + \angle \:ABC = 180$

$\tt \:  \longrightarrow \: \angle \: ABC \: + 80 = 180$

$\tt \:  \longrightarrow \: \angle \: ABC \: = 180 - 80$

$\tt \:  \longrightarrow \: \angle \: ABC \: = 100$

☆ In triangle ️ ABF

☆ Using angle sum property of triangle,

$\tt \:  \longrightarrow \: \angle \:FAB + \angle \:FBA +\angle \: AFB = 180$

$\tt \:  \longrightarrow \: \angle \:AFB + 50 + 100 = 180$

$\tt \:  \longrightarrow \: \angle \:AFB + 150 = 180$

$\tt \:  \longrightarrow \: \angle \:AFB = 180 - 150$

$\tt \:  \longrightarrow \: \angle \:AFB = 30$

$\bf\implies \: \boxed{ \red{\bf \:  \angle \:CFD \: = \: 30 \degree}}$