(B) inthe figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also Prove that OE is parallel to BD.

Solution:

(a) Join DB, CA and CB. ∠ADC = 118° (given) and ∠ADB = 90°

(Angles in a semi-circle)

∠BDC = ∠ADC – ∠ADB

= 118⁰ – 90^o = 28^o

∠ABCD is a cyclic quadrilateral)

∠ADC + ∠ABC = 180^o

118^o + ∠ABC = 180^o

∠ABC = 180^o – 118^o = 62^o

But in ∆AEB

∠AEB = 90^o

(Angles in a semi-circle)

∠EAB = ∠ABE (AE = BE)

∠EAB + ∠ABE = 90^o

∠EAB = 90^o × ½ = 45^o

∠CBE = ∠ABC + ∠ABE

= 62^o + 45^o = 107^o

But AEBD is a cyclic quadrilateral

∠CAE + ∠CBE = 180^o

∠CAE + 107^o = 180^o

∠CAE = 180^o – 107^o = 73^o

(b) AB is the diameter of semi-circle ABCDE

With center O.AE = ED and ∠BCD = 140^o

In cyclic quadrilateral EBCD.

(i) ∠BCD + ∠BED = 180o

140o + ∠BED = 180o

∠BED = 180o – 140^o = 40⁰

But ∠AED = 90^o

(Angles in a semi circle)

∠AED = ∠AEB + ∠BED

= 90^o + 40^o = 130^o

(ii) Now in cyclic quadrilateral AEDB

∠AED + ∠DBA = 180^o

130^o + ∠DBA =180^o

∠BDA = 180^o – 130^o = 50^o

Chord AE = ED (given)

∠DBE = ∠EBA

But ∠DBE + ∠EBA = 50^o

DBE + ∠DBE = 50^o

2∠DBE = 50^o

∠DBE = 25^o or ∠EBD = 25^o

In ∆OEB,OE = OB

(raddi of the same circle)

∠OEB = ∠EBO = ∠DBE

But these are ultimate angles

OE ∥ BD