Question

6. In figure, ABCD is a cyclic quadrilateral in which AB is extended till F and BE || DC. If angleFBE=20° and DAB = 95°, then find angleADC.​

Answer 1

Answer:

Sum of opposite angles of a cyclic quadrilateral is 180°

Sum of opposite angles of a cyclic quadrilateral is 180°∴ ∠DAB + ∠BCD = 180°

Sum of opposite angles of a cyclic quadrilateral is 180°∴ ∠DAB + ∠BCD = 180° ⇒ 95° + ∠BCD = 180°

Sum of opposite angles of a cyclic quadrilateral is 180°∴ ∠DAB + ∠BCD = 180° ⇒ 95° + ∠BCD = 180° ⇒ ∠BCD = 180° - 95° = 85°

Sum of opposite angles of a cyclic quadrilateral is 180°∴ ∠DAB + ∠BCD = 180° ⇒ 95° + ∠BCD = 180° ⇒ ∠BCD = 180° - 95° = 85°∵ BE || DC

Sum of opposite angles of a cyclic quadrilateral is 180°∴ ∠DAB + ∠BCD = 180° ⇒ 95° + ∠BCD = 180° ⇒ ∠BCD = 180° - 95° = 85°∵ BE || DC∴ ∠CBE = ∠BCD = 85° [alternate interior angles]

Sum of opposite angles of a cyclic quadrilateral is 180°∴ ∠DAB + ∠BCD = 180° ⇒ 95° + ∠BCD = 180° ⇒ ∠BCD = 180° - 95° = 85°∵ BE || DC∴ ∠CBE = ∠BCD = 85° [alternate interior angles]∴ ∠CBF = ∠CBE + ∠FBE

Sum of opposite angles of a cyclic quadrilateral is 180°∴ ∠DAB + ∠BCD = 180° ⇒ 95° + ∠BCD = 180° ⇒ ∠BCD = 180° - 95° = 85°∵ BE || DC∴ ∠CBE = ∠BCD = 85° [alternate interior angles]∴ ∠CBF = ∠CBE + ∠FBE = 85° + 20° = 105°

Sum of opposite angles of a cyclic quadrilateral is 180°∴ ∠DAB + ∠BCD = 180° ⇒ 95° + ∠BCD = 180° ⇒ ∠BCD = 180° - 95° = 85°∵ BE || DC∴ ∠CBE = ∠BCD = 85° [alternate interior angles]∴ ∠CBF = ∠CBE + ∠FBE = 85° + 20° = 105° Now, ∠ABC + ∠CBF = 180° [linear pair]

Sum of opposite angles of a cyclic quadrilateral is 180°∴ ∠DAB + ∠BCD = 180° ⇒ 95° + ∠BCD = 180° ⇒ ∠BCD = 180° - 95° = 85°∵ BE || DC∴ ∠CBE = ∠BCD = 85° [alternate interior angles]∴ ∠CBF = ∠CBE + ∠FBE = 85° + 20° = 105° Now, ∠ABC + ∠CBF = 180° [linear pair]and ∠ABC + ∠ADC = 180° [opposite angles of cyclic quad]

Sum of opposite angles of a cyclic quadrilateral is 180°∴ ∠DAB + ∠BCD = 180° ⇒ 95° + ∠BCD = 180° ⇒ ∠BCD = 180° - 95° = 85°∵ BE || DC∴ ∠CBE = ∠BCD = 85° [alternate interior angles]∴ ∠CBF = ∠CBE + ∠FBE = 85° + 20° = 105° Now, ∠ABC + ∠CBF = 180° [linear pair]and ∠ABC + ∠ADC = 180° [opposite angles of cyclic quad]Thus, ∠ABC + ∠ADC = ∠ABC + ∠CBF

Sum of opposite angles of a cyclic quadrilateral is 180°∴ ∠DAB + ∠BCD = 180° ⇒ 95° + ∠BCD = 180° ⇒ ∠BCD = 180° - 95° = 85°∵ BE || DC∴ ∠CBE = ∠BCD = 85° [alternate interior angles]∴ ∠CBF = ∠CBE + ∠FBE = 85° + 20° = 105° Now, ∠ABC + ∠CBF = 180° [linear pair]and ∠ABC + ∠ADC = 180° [opposite angles of cyclic quad]Thus, ∠ABC + ∠ADC = ∠ABC + ∠CBF ⇒ ∠ADC = ∠CBF

Sum of opposite angles of a cyclic quadrilateral is 180°∴ ∠DAB + ∠BCD = 180° ⇒ 95° + ∠BCD = 180° ⇒ ∠BCD = 180° - 95° = 85°∵ BE || DC∴ ∠CBE = ∠BCD = 85° [alternate interior angles]∴ ∠CBF = ∠CBE + ∠FBE = 85° + 20° = 105° Now, ∠ABC + ∠CBF = 180° [linear pair]and ∠ABC + ∠ADC = 180° [opposite angles of cyclic quad]Thus, ∠ABC + ∠ADC = ∠ABC + ∠CBF ⇒ ∠ADC = ∠CBF ⇒ ∠ADC = 105° [∵∠CBF = 105°]

Answer 2

Answer:

∠ ADC = 105°

Explanation:

Sol. Since ABCD is a cyclic quadrilateral.

∴∠  BAD + ∠  BCD = 180°

⇒  95° + ∠  BCD = 180° ⇒ ∠  BCD = 85°

Since, BE || DC therefore, ∠BCD = ∠  CBE ⇒ ∠  CBE = 85° [Alternate angles]

Now, ∠  BCF = ∠  CBE + ∠  EBF

= 85° + 20° = 105°

Now, since exterior angle formed by producing a side of a cyclic quadrilateral, is equal to the interior op­posite angle.

∠  ADC = ∠  BCF = 105°.

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