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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°

Mathematics
RS Aggarwal (2017)
Standard IX

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