Spot equal sides to build isosceles triangles

4.MD.C.74.G.A.2 · take · grade 4

Archetype: Isosceles and Equilateral Angle Chaining · step in a 6-type progression

Quadrilateral $ABCD$ is a square, and triangle $CED$ is an equilateral triangle. Find the measure of $\angle a$ .

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Understand

ABCD is a square and CED is an equilateral triangle built on the outside of side DC, with E pointing right. Segment BE crosses side DC at F. I must find angle a, the angle at F.

Givens

ABCD is a square, so all sides are equal and every corner angle is 90 degrees
Triangle CED is equilateral on side DC, so DC = CE = DE and each of its angles is 60 degrees
Side BC of the square equals side CE of the triangle (both equal the square's side)
Segment BE meets side DC at point F; angle a is the angle at F

Unknowns

The measure of angle a at point F

Constraints

An isosceles triangle has two equal base angles
Angles in a triangle sum to 180 degrees
Square corner angles are 90 degrees and equilateral angles are 60 degrees

Plan

#7 Identify Subproblems · also uses: #1 Draw a Diagram

The trick is to spot that side BC (square) and side CE (equilateral triangle) are equal, so triangle BCE is isosceles. I find its base angles, then use triangle BFC (with the square's right angle at C) to reach angle a at F.

Execute

#7 Identify Subproblems 4.MD.C.7

At corner C, the square contributes angle BCD = 90 degrees and the equilateral triangle contributes angle DCE = 60 degrees on the outside. Together angle BCE = 90 + 60 = 150 degrees.

\angle BCE = 90^\circ + 60^\circ = 150^\circ

Angles that meet at one point add up, so the square angle plus the triangle angle gives the whole angle at C.

#7 Identify Subproblems 4.MD.C.7

Side BC of the square equals side CE of the equilateral triangle, so triangle BCE is isosceles with BC = CE. Its two base angles are equal: angle CBE = angle CEB = (180 - 150) / 2 = 15 degrees.

\angle CBE = \angle CEB = \dfrac{180^\circ - 150^\circ}{2} = 15^\circ

Equal sides give equal base angles, and the three angles still total 180 degrees.

#7 Identify Subproblems 4.MD.C.7

F lies on side DC, so in triangle BFC the angle at C (angle FCB) is the square's corner, 90 degrees, and angle FBC = angle CBE = 15 degrees. So angle BFC = 180 - 90 - 15 = 75 degrees. Angle a at F is the angle on the upper-left, the straight-line partner of angle BFC: angle a = 180 - 75 = 105 degrees.

\angle BFC = 180^\circ - 90^\circ - 15^\circ = 75^\circ,\quad \angle a = 180^\circ - 75^\circ = 105^\circ

Once two angles of triangle BFC are known the third follows, and angle a is its straight-line partner along DC.

Answer: 105 degrees

Review

Angle a = 105 degrees is obtuse, which fits the wide angle the crossing line BE makes with side DC on the upper side, while its partner angle BFC = 75 degrees is acute; together they make 180 degrees along the straight side DC, so the result is consistent.

Place the square on a coordinate grid (tool 1), compute the position of E and F, and measure angle a directly to confirm the 105-degree result from the isosceles-triangle reasoning.

Standards · min grade 4

4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding the square and triangle angles at C, finding isosceles base angles, and chaining through triangle BFC to angle a.
4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using square and equilateral-triangle side and angle properties to spot the isosceles triangle BCE.

💡 This only needs Grade 4 angle-adding plus spotting that a square side and a triangle side are equal!