Question

∆ ABC is an equilateral triangle. D is any point on BC. Another equilateral ∆BDE is constructed outside the ∆ABC. AD and CE are joined. Show that AD = CE . just say how to procede

Answer 1

Answer:

Try to prove Triangles ABD & BCE are congruent.(SAS)

Step-by-step explanation:

AB=BC

BD=BE

<ABD =<CBE

Answer 2

Given :  ∆ ABC is an equilateral triangle. D is any point on BC. Another equilateral ∆BDE is constructed outside the ∆ABC.  

To Find : Show that AD = CE .

Solution:

AB = BC = AC  = x

BD = BE = DE = y

∠ABD = 60°   ∵ ∠ABC = 60°

∠CBE = 60°    ∵ ∠DBE = 60°  

Comparing  ΔABD & ΔCBE

AB  = BC   =   x

∠ABD = ∠CBE  = 60°

BD = BE    = y

=> ΔABD ≅ ΔCBE   ( SAS crieteria)

=> AD = CE

Another method : Cosine rule

AD² = AB² + BD² - 2AB . BD Cos∠ABD

=> AD² = x² + y² - 2xyCos(60°)

=> AD² = x² + y² - xy    ∵ Cos(60°) = 1/2

CE² = CD² + DE² - 2CD . DE Cos∠CDE

CD = BC - BD = x - y    , DE = y

∠CDE = 180° - ∠BDE = 180° - 60° = 120°

=> CE² =  (x-y)² + y²- 2(x-y)yCos(120°)

=> CE² = x² + y² - 2xy  + y²  + xy - y²      ∵ Cos(120°) = -1/2

=> CE² =  x² + y² - xy

AD² =   CE² =  x² + y² - xy

=> AD = CE

QED

Hence proved

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