Mathematics
In the figure (i) given below, ED and BC are two parallel chords of the circle and ABE, ACD are two st. lines. Prove that AED is an isosceles triangle.
Answer
BEDC is a cyclic quadrilateral as all of its vertices lie on the circumference of the circle.
∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.
∠ABC = ∠D ….(i)
From figure, as ED is parallel to BC, ∠ABC and ∠E are corresponding angles,
∴ ∠ABC = ∠E ….(ii)
From (i) and (ii)
∠D = ∠E
In △AED,
∠D = ∠E,
∴ AE = AD. (As sides opposite to equal angles are equal)
Hence, proved that △AED is an isosceles triangle.
Related Questions
In the figure (ii) given below, O is the center of the circle. If ∠BAD = 75° and BC = CD, find:
(i) ∠BOD
(ii) ∠BCD
(iii) ∠BOC
(iv) ∠OBD
In the adjoining figure, ABC is an isosceles triangle in which AB = AC and circle passing through B and C intersects sides AB and AC at points D and E. Prove that DE || BC.
In the figure (ii) given below, SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = RS.
In the adjoining figure, O is the center and AOE is the diameter of the semicircle ABCDE. If AB = BC and ∠AEB = 50°, find :
(i) ∠CBE
(ii) ∠CDE
(iii) ∠AOB.
Prove that OB is parallel to EC.
