In figure, ABCD is a square and ∠DEC is an equilateral triangle. Prove that
(i) ∆ADE ≅ ∆BCE
(ii) AE = BE
(iii) ∠DAE = 15°
Given: ABC is an isosceles triangle with
AB = AC.
AP ⊥ BC
To Prove: ∠B = ∠C
Proof: In ∆ABC,
∵ AB = AC | Given
∴ ∠ABC = ∠ACB ...(1)
| Angles opposite to equal sides of a triangle are equal
Now, in ∆APB and ∆APC,
AB = AC | Given
∠ABP = ∠ACP | From (1)
∠APB = ∠APC (= 90°) | Given
∴ ∆APB ≅ ∆APC | AAS congruence rule
∴ ∠ABP = ∠ACP | CPCT
⇒ ∠B = ∠C