Question

4. In the adjoining figure, ABCD is a square and
AEDC is an equilateral triangle. Prove that
(i) AE =BE, (ii) ZDAE = 15° ​

Answer 1

${\sf{\underline{\overline{Question:-}}}}$

In the adjoining figure, ABCD is a square and

AEDC is an equilateral triangle. Prove that

(i) AE =BE,

(ii) ZDAE = 15°

${\sf{\underline{\overline{AnsWeR:-}}}}$

${\sf{\green{\underline{\underline{Given:}}}}}$

• ABCD is Square
• EDC is an equilateral triangle

• AD= BC
• AD= BCDE= CE

To Prove:-

• AE= BE
• |DAE|= 15°

Construction => join

• A to E
• B to E

Prove =>

• In ∆ADE and ∆BCE
• AD= BC (given)
• | ADE = | BCE
• DE = CE (Given)

Option:-

• AE =BE,

Answer 2

Answer:

Question:−

In the adjoining figure, ABCD is a square and

AEDC is an equilateral triangle. Prove that

Given:

(i) AE =BE,

(ii) ZDAE = ABCD is Square

EDC is an equilateral triangle

AD= BC

AD= BCDE= CE

To Prove:-

AE= BE

|DAE|= 15°

Construction => join

A to E

B to E

Prove =>

In ∆ADE and ∆BCE

AD= BC (given)

| ADE = | BCE

DE = CE (Given)

Option:-

AE =BE,