Mathematics
In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side BC ∥ AE. If ∠BAC = 50°, find giving reasons :
(i) ∠ACB
(ii) ∠EDC
(iii) ∠BEC
Hence prove that BE is also a diameter.
Circles
10 Likes
Answer
(i) ∠ABC = 90° [Angle in semicircle]
In ∆ABC,
⇒ ∠ABC + ∠BAC + ∠ACB = 180° [Angle sum property of triangle]
⇒ 90° + 50° + ∠ACB = 180°
⇒ 140° + ∠ACB = 180°
⇒ ∠ACB = 40°
Hence, ∠ACB = 40°.
(ii) ∠EAC = ∠ACB = 40° [Alternate angles, AC transversal to parallel lines AE and BC]
∠EAC + ∠EDC = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]
40° + ∠EDC = 180°
∠EDC = 140°
Hence, ∠EDC = 140°.
(iii) ∠EBC + ∠EDC = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]
140° + ∠EBC = 180°
∠EBC = 40°
In ∆EBC,
⇒ ∠BEC + ∠ECB + ∠EBC = 180° [Angle sum property of triangle]
⇒ ∠BEC + 90° + 40° = 180°
⇒ ∠BEC = 180° - 130°
⇒ ∠BEC = 50°
In ∆EAB,
⇒ ∠EAB = ∠EAC + ∠BAC
= 40° + 50° = 90°
We know that, if an angle of a triangle in a circle is 90° then, the hypotenuse must be the diameter of the circle.
Hence, ∠BEC = 50° and BE is the diameter of the circle.
Answered By
4 Likes
Related Questions
In the given figure, I is the incentre of Δ ABC. Here AI produced meets the circumcircle of Δ ABC at D. If ∠ABC = 55° and ∠ACB = 65°, calculate :
(i) ∠BCD
(ii) ∠CBD
(iii) ∠DCI
(iv) ∠BIC
In the given figure, ∠DBC = 58° and BD is a diameter of the circle. Calculate :
(i) ∠BDC
(ii) ∠BEC
(iii) ∠BAC
In the given figure, O is the centre of the circle and AB is a diameter. If AC = BD and ∠AOC = 72°, find:
(i) ∠ABC
(ii) ∠BAD
(iii) ∠ABD
In the given figure, A, C, B, D are points on the circle with centre O. Given ∠ABC = 62°. Find :
(i) ∠ADC
(ii) ∠CAB
