Question

(
b) In the figure (ii) given below, AB is the diameter of the semicircle ABCDE with
centre O. IF AE = ED and ZBCD = 140°, find /_AED and /_EBD. Also prove that OE is parallel to BD.
​

Answer 1

AB is the diameter of semi-circle ABCDE

With center O.AE = ED and ∠BCD = 140o

In cyclic quadrilateral EBCD.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 55

(i) ∠BCD + ∠BED = 180o

140o + ∠BED = 180o

∠BED = 180o – 140o = 400

But ∠AED = 90o

(Angles in a semi circle)

∠AED = ∠AEB + ∠BED

= 90o + 40o = 130o

(ii) Now in cyclic quadrilateral AEDB

∠AED + ∠DBA = 180o

130o + ∠DBA =180o

∠BDA = 180o – 130o = 50o

Chord AE = ED (given)

∠DBE = ∠EBA

But ∠DBE + ∠EBA = 50o

DBE + ∠DBE = 50o

2∠DBE = 50o

∠DBE = 25o or ∠EBD = 25o

In ∆OEB,OE = OB

(raddi of the same circle)

∠OEB = ∠EBO = ∠DBE

But these are ultimate angles

OE ∥ BD