Question

In the adjoining figure, ABCD is a square and ADE is an equilateral triangle. Prove that `∠ACE = 1/2 ∠ DAE`.

Answer 1

Given:

• ABCD is a square.
• ADE is an equilateral triangle with AD as one side of the equilateral triangle.

To Prove:

• `∠ACE = 1/2 ∠DAE`.

Proof [Step-wise]:

1. Place the square in a coordinate plane for a clean calculation.

Let the side length be 1 and take A = (0, 1), B = (0, 0), C = (1, 0), D = (1, 1).

2. Because triangle ADE is equilateral with base AD of length 1 from A(0, 1) to D(1, 1), the third vertex E lies above the midpoint of AD at height `h = (sqrt(3))/2`.

Thus, `E = (1/2, 1 + sqrt(3)/2)`.

3. Compute the direction angle of CA and CE measured from the positive x-axis.

Vector CA = A – C

= (0 – 1, 1 – 0)

= (–1, 1)

The direction angle of CA is 135°.

Vector CE = E − C

= `(1/2 - 1, 1 + sqrt(3)/2 - 0)`

= `(-1/2, 1 + sqrt(3)/2)`

4. Find the direction angle of CE.

Its slope is slope (CE)

= `(1 + sqrt(3)/2)/(-1/2)`

= `-2(1 + sqrt(3)/2)`

= `-(2 + sqrt(3))`

Note the identity `tan 75° = 2 + sqrt(3)`.

Since tan (45° + 30°) 

= `(1 + 1/sqrt(3))/(1 - 1/sqrt(3))`

= `2 + sqrt(3)`

Therefore, arc `tan (2 + sqrt(3)) = 75°`.

So, arc `tan -(2 + sqrt(3)) = -75^circ` and the direction angle of CE in 0° – 360° is 180° – 75° = 105°.

5. Thus, the angle at C between CA and CE is ∠ACE

= Direction (CA) – Direction (CE)

= 135° – 105°

= 30°

6. In triangle ADE (equilateral), we have ∠DAE = 60°.

Hence, ∠ACE = 30° 

= `1/2 xx 60^circ`

= `1/2 ∠DAE`

`∠ACE = 1/2 ∠DAE`, as required.