Question

Points A, B, C and D lie on a circle with AB = 4
and BC = 2. AC is a diameter and angle ABD =
angle CBD. What is BD?
(A) 2/2
(B) 3√2
(C) √10
(D) 2/3
​

Answer 1

Given :  Points A, B, C and D lie on a circle with  AB = 4   and BC = 2 .

AC is a diameter

∠ABD = ∠CBD

To Find : BD

(A) 2/2

(B) 3√2

(C) √10

(D) 2/3

Solution:

AB = 4

BC = 2

AC is diameter hence ∠ABC = 90°

=> AC² = AB² + BC²

=> AC² = 4² + 2²

=> AC = 2√5

∠ABD = ∠CBD  

Hence BD is angle bisector

=> AB /BC = AE/CE

=> 4/2  = AE/CE

=> AE = 2CE

AE + CE = AC = 2√5

=> 3CE = 2√5

=> CE = 2√5 / 3

   AE = 2 CE = 4 √5 / 3

ΔABE  & ΔDBC

∠BAE = ∠BDC      (∵ ∠BAC = ∠BDC by same arc BC )

∠ABE = ∠DBC    

=> ΔABE ≈ ΔDBC

=> AB/BD  = BE/ BC

=> 4/BD  = BE/2

=> BD  BE = 8

BE  DE = AE   CE  ( as two chord  intersects at E )

=> BE  DE = ( 4 √5 / 3) (2√5 / 3)

=> BE   DE = 40/9

BD  BE = 8

BD = BE + DE

=> ( BE + DE )  BE = 8

=> BE² + BE DE = 8

=> BE² + 40/9 = 8

=> BE² = 32/9

=> BE = 4√2/3

BD  BE = 8

=> BD   4√2/3 =  8

=> BD = 3√2

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