In the adjoining figure, ABCD is a square and ΔEDC is an equilateral triangle. Prove that (i) AE=BE, (ii) ∠DAE=15∘.
Given : ABCD is a square and EDC is an equilateral triangle. AD=BC=CD=DE=CE
To Prove : i) AE=BE
ii) ∠DAE=15∘
Construction : Join A to E and B to E.
Proof :
i) In △ADE and △BCE, we have
AD=BC(given)
∠ADE=∠BCE(=90∘+60∘)
DE=CE(given)
Therefore, △ADE≅△BCE (By SAS rule)
AE=BE (CPCT)
ii) ∠DAE+∠ADE+∠DEA=180∘ [Angle sum property]
⇒150∘+∠DAE+∠DEA=180∘
⇒∠DAE+∠DEA=180∘−150∘ [AD=ED, angle opposite to equal sides are equal, ∵∠DAE=∠DEA ]
⇒2∠DAE=30∘
⇒∠DAE=15∘