Question

\(ABCD\) is a square (vertices in order \(A\), \(B\), \(C\), \(D\)) and \(BCE\) is an equilateral triangle drawn on side \(BC\), with vertex \(E\) lying outside the square. What is the measure of angle \(DEC\)?

• A. 15°
• B. 30°
• C. 20°
• D. 45°

Hint:

Add angle DCB (90°) and angle BCE (60°) to get angle DCE, then use isosceles triangle DCE.

Answer 1

The Correct Option is A

Step 1: Identify the known angles at point C.
\(ABCD\) is a square, so every interior angle measures \(90^\circ\); in particular \(\angle DCB = 90^\circ\). \(BCE\) is equilateral, so every angle of it measures \(60^\circ\); in particular \(\angle BCE = 60^\circ\).

Step 2: Find angle DCE.
Since \(E\) lies outside the square, on the far side of \(BC\), the angle \(\angle DCE\) is made up of \(\angle DCB\) and \(\angle BCE\) placed next to each other around point \(C\):
\[ \angle DCE = 90^\circ + 60^\circ = 150^\circ \]

Step 3: Use the isosceles triangle DCE.
\(DC\) is a side of the square and \(CE\) is a side of the equilateral triangle. Since \(BC\) is shared by both shapes, \(DC = BC = CE\). So triangle \(DCE\) is isosceles with \(DC = CE\), meaning the base angles opposite these equal sides are equal: \(\angle CDE = \angle CED\). The angles of the triangle sum to \(180^\circ\):
\[ \angle CDE + \angle CED = 180^\circ - 150^\circ = 30^\circ \]
Since these two angles are equal:
\[ \angle DEC = \frac{30^\circ}{2} = 15^\circ \]

Final Answer:
Angle DEC measures \(15^\circ\), so option A is correct. \[ \boxed{\angle DEC = 15^\circ} \]