In figure, ABCD is a square and ∠DEC is an equilateral triangle. Prove that
(i) ∆ADE ≅ ∆BCE
(ii) AE = BE
(iii) ∠DAE = 15°
Given: In an isosceles triangle ABC with AB = AC, BD and CE are two medians.
To Prove: BD = CE
Proof: In ∆ABC,
∵ AB = AC
∴ ∠BC = ∠ACB ...(1)
| Angles opposite to equal sides of a triangle are equal
Also,
| Halves of equals are equal ⇒ BE = CD ...(2)
| ∵ BD and CE are two medians
Now, in ∆BDC and ∆CEB,
∠BCD = ∠CBE | From (1)
BE = CD | From (2)
BC = CB | Common
∴ ∆BDC ≅ ∆CEB
| SAS congruence rule
∴ BD = CE | CPCT