In the given figure, ABCD is a cyclic quadrilateral in which DC is produced to E and CF is drawn parallel to AB such that ∠ADC = 95° and ∠ECF = 20°.
Question:

In the given figure, ABCD is a cyclic quadrilateral in which DC is produced to E and CF is drawn parallel to AB such that ADC = 95° and ∠ECF = 20°. Then, ∠BAD = ?
(a) 95°

(b) 85°
(c) 105°
(d) 75°

 

Solution:

(c) 105°
We have:
∠ABC + ∠ADC = 180°
⇒ ∠ABC + 95° = 180°
∠ABC = (180° – 95°) = 85°
Now,
 CF || AB and CB is the transversal.
∴ ∠BCF = ∠ABC = 85°     (Alternate interior angles)
⇒ ∠BCE = (85° + 20°) = 105°
⇒ ∠DCB = (180° – 105°) = 75°
Now, ∠BAD + ∠BCD = 180°

⇒ ∠BAD + 75° = 180°
⇒ ∠BAD = (180° – 75°) = 105°

 

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