Question

If ABCD is a square and BCE is an equilateral triangle, what is the measure of $\angle DEC$?

• A. $15^\circ$
• B. $30^\circ$
• C. $20^\circ$
• D. $45^\circ$

Hint:

Combine known angles from square and equilateral triangle properties.

Answer 1

The Correct Option is A

To find the measure of $\angle DEC$, let's dissect the given problem:

• Square ABCD: In a square, all angles are $90^\circ$. 
• Equilateral Triangle BCE: In an equilateral triangle, all angles are $60^\circ$.

Knowing these facts, we can solve for $\angle DEC$:

1. $\angle DAB = 90^\circ$ because ABCD is a square.
2. $\angle CBE = 60^\circ$ because BCE is an equilateral triangle.
3. Since $\angle ABC = 90^\circ$ (part of square ABCD), and knowing $\angle CBE = 60^\circ$, we determine $\angle EBC = 30^\circ$ (as part of $\angle ABC$), because $90^\circ - 60^\circ = 30^\circ$.
4. In triangle CDE, $\angle DCE = \angle EBC = 30^\circ$ (alternate angles) because DE and AC are both parallel lines.
5. In triangle CDE, the sum of angles is $180^\circ$: $\angle DCE + \angle DEC + \angle CED = 180^\circ$.
6. Since $\angle CDE = 90^\circ$ (part of square ABCD), we substitute: $30^\circ + \angle DEC + 90^\circ = 180^\circ$.
7. Simplify the equation: $\angle DEC = 180^\circ - 120^\circ = 60^\circ$.

Therefore, the measure of $\angle DEC$ is $15^\circ$.