Question

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

Answer 1

Question :- ∆ABC is an equilateral triangle. D is any point on BC. Another equilateral ∆ is constructed outside the ∆. AD and CE are joined. Show that AD = CE ?

To Show :- AD = CE

Diagram :- Refer to image.

Solution :-

From diagram , we have two Equaliteral ∆'s :- ∆ABC and ∆EBD.

in ∆ABC :-

→ ∠ABC = ∠BAC = ∠ACB = 60°

→ AB = BC = CA

Similarly, in ∆EBD :-

→ ∠BDE = ∠BED = ∠DBE = 60°

→ AB = BC = CA

In Equaliteral ∆ all sides are Equal in Length and Measure of Each three angles is 60° .

Now, in ∆ABD and ∆CBE ,

→ AB = CB (side of Equaliteral ∆ABC).

→ ∠ABD = ∠CBE. (60° Each) .

→ BD = BE. ( side of Equaliteral ∆EBD).

Therefore,

→ ∆ABD ≅ ∆CBE .

Hence,

→ AD = CE. (By CPCT).