In triangle ABC, AC = 3AB. If AD bisects angle A with D lying on BC and E is the foot of the
perpendicular from C to AD, then area of AABD/area of ACDE
is equal to...........
Answers
Given : in Δ ABC AC = 3AB
AD bisects angle A with D lying on BC and E is the foot of the perpendicular from C to AD
To find : Area of ΔABD / Area of Δ CDE
Solution:
in Δ ABC
AD bisects angle A with D lying on BC
AB/AC = BD/CD
AC = 3AB
=> CD = 3 BD
Let say AB = m then AC = 3m
BD = n then CD = 3n
Expand AB and CE to intersect at F
ΔAEF ≅ ΔAEC ( AD is angle bisector and AD common and AE ⊥ FC)
=> AF = AC = 3m and FE = EC
=> BF = AF - AB = 3m - m = 2m
as ΔAEF ≅ ΔAEC
similarly ΔDEF ≅ ΔDEC
let say Area of ΔBDF = 2A
then area of ΔCDF = 3 * 2A = 6A
(as CD/BD = 3n/n = 3)
Area of Δ DEF = Area of Δ CDE = area of ΔCDF/ 2 = 6A/2 = 3A
Area of ΔBDF = 2A
=> Area of ΔABD = (1/2)2A = A as AB/BF = m/2m = 1/2
Area of ΔABD / Area of Δ CDE = A/ 3A
=> Area of ΔABD / Area of Δ CDE = 1/ 3
Area of ΔABD / Area of Δ CDE = 1/ 3
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