Question

in the figure below, $\triangle abc$ is equilateral, $\triangle def$ is isosceles with $de = df$, and $\overline{ef}$ is parallel to $\overline{ac}$. if $m\angle dfc = 85^\circ$, what is the measure of $\angle edf$? select one answer a $70^\circ$ b $85^\circ$ c $90^\circ$ d $110^\circ$

Explanation:

Step1: Find angle in equilateral triangle

$\triangle ABC$ is equilateral, so $\angle C = 60^\circ$.

Step2: Find $\angle DFC$'s relation to $\triangle DFC$

In $\triangle DFC$, we know $\angle DFC = 85^\circ$ and $\angle C = 60^\circ$. Wait, no, first, since $EF \parallel AC$, and $\triangle ABC$ is equilateral, $\angle BAC = \angle ABC = \angle ACB = 60^\circ$. Also, $\triangle DEF$ is isosceles with $DE = DF$, so $\angle DEF = \angle DFE$. Wait, maybe better: since $EF \parallel AC$, $\angle DFE = \angle FDC$? No, wait, $\angle DFC = 85^\circ$, and $\angle ACB = 60^\circ$. Wait, no, let's correct. $\triangle ABC$ is equilateral, so $\angle C = 60^\circ$. In $\triangle DFC$, angles sum to $180^\circ$, but wait, no, actually, since $EF \parallel AC$, $\angle DFE = \angle FDC$? Wait, no, the key is that $\triangle DEF$ is isosceles with $DE = DF$, so base angles $\angle DEF = \angle DFE$. Also, since $EF \parallel AC$, $\angle DFE = \angle FDC$? Wait, no, $\angle DFC = 85^\circ$, and $\angle ACB = 60^\circ$. Wait, maybe first find $\angle FDC$. Wait, no, $\angle DFC = 85^\circ$, $\angle C = 60^\circ$, so in $\triangle DFC$, $\angle FDC = 180^\circ - 85^\circ - 60^\circ = 35^\circ$? No, that can't be. Wait, no, $EF \parallel AC$, so $\angle DFE = \angle FDC$? Wait, no, maybe $\angle DFE = \angle FDC$ because of alternate interior angles. Wait, no, let's start over.

$\triangle ABC$ is equilateral, so all angles are $60^\circ$. $EF \parallel AC$, so $\angle DFE = \angle FDC$ (alternate interior angles). Wait, no, $\angle DFC = 85^\circ$, so $\angle DFE = 180^\circ - 85^\circ = 95^\circ$? No, that's supplementary. Wait, no, $E - F - C$? No, the diagram: points $E$ on $AB$, $F$ on $BC$, $D$ on $AC$. So $EF$ is parallel to $AC$, so $\angle BEF = \angle BAC = 60^\circ$, $\angle BFE = \angle BCA = 60^\circ$, so $\triangle BEF$ is equilateral? Wait, maybe not. $\triangle DEF$ is isosceles with $DE = DF$, so $\angle DEF = \angle DFE$. Now, $\angle DFC = 85^\circ$, so $\angle DFE = 180^\circ - 85^\circ = 95^\circ$? No, that's if $F$ is on $BC$, so $\angle DFC$ and $\angle DFE$ are supplementary. So $\angle DFE = 180 - 85 = 95^\circ$? But $\triangle DEF$ is isosceles with $DE = DF$, so $\angle DEF = \angle DFE = 95^\circ$? Then $\angle EDF = 180 - 95 - 95 = -10^\circ$? That's impossible. So I must have messed up.

Wait, no, $\triangle ABC$ is equilateral, so $\angle C = 60^\circ$. $\angle DFC = 85^\circ$, so in $\triangle DFC$, $\angle FDC = 180 - 85 - 60 = 35^\circ$. Then, since $EF \parallel AC$, $\angle EDF = 180 - 2 \times 35$? No, wait, $\triangle DEF$ is isosceles with $DE = DF$, so $\angle DEF = \angle DFE$. Also, $EF \parallel AC$, so $\angle DFE = \angle FDC = 35^\circ$? Then $\angle EDF = 180 - 35 - 35 = 110^\circ$? Wait, no, that doesn't match. Wait, maybe $\angle DFE = 85^\circ$? Wait, the problem says $m\angle DFC = 85^\circ$, so $\angle DFC = 85^\circ$, and since $EF \parallel AC$, $\angle DFE = \angle FDC$? No, maybe $\angle DFE = 85^\circ$? Wait, no, $\angle DFC$ and $\angle DFE$ are adjacent angles on a straight line? No, $F$ is on $BC$, so $E - F - C$? No, $E$ is on $AB$, $F$ is on $BC$, $D$ is on $AC$. So $EF$ is parallel to $AC$, so $\angle DFE = \angle FDC$ (alternate interior angles). $\angle DFC = 85^\circ$, $\angle C = 60^\circ$, so $\angle FDC = 180 - 85 - 60 = 35^\circ$. Then $\angle DFE = 35^\circ$? But $\triangle DEF$ is isosceles with $DE = DF$, so $\angle DEF = \angle DFE = 35^\circ$, so $\angle EDF = 180 - 35 - 35 = 110^\circ$? Wait, that gives $110^\circ$, which is option D? Wait, no, let's check again.

Wait, $\triangle ABC$ is equilateral, so $\angle C = 60^\circ$. $\angle DFC = 85^\circ$, so in $\triangle DFC$, $\angle FDC = 180 - 85 - 60 = 35^\circ$. Then, since $EF \parallel AC$, $\angle DFE = \angle FDC = 35^\circ$ (alternate interior angles). $\triangle DEF$ is isosceles with $DE = DF$, so $\angle DEF = \angle DFE = 35^\circ$. Therefore, $\angle EDF = 180 - 35 - 35 = 110^\circ$. Wait, but that seems high. Wait, no, maybe $\angle DFE = 85^\circ$? Wait, $\angle DFC = 85^\circ$, so $\angle DFE = 180 - 85 = 95^\circ$ (supplementary). Then $\triangle DEF$ with $DE = DF$, so $\angle DEF = \angle DFE = 95^\circ$, then $\angle EDF = 180 - 95 - 95 = -10^\circ$, which is impossible. So my first approach was wrong.

Wait, maybe $\angle DFE = \angle DFC$? No, $EF \parallel AC$, so $\angle DFE = \angle FDC$ (alternate interior angles). Wait, $\angle DFC = 85^\circ$, $\angle C = 60^\circ$, so $\angle FDC = 180 - 85 - 60 = 35^\circ$. Then $\angle DFE = 35^\circ$, so $\triangle DEF$ has $\angle DFE = 35^\circ$, $\angle DEF = 35^\circ$, so $\angle EDF = 180 - 35 - 35 = 110^\circ$. Yes, that works. So the answer is $70^\circ$? Wait, no, 180 - 35 - 35 is 110? Wait, 35 + 35 is 70, 180 - 70 is 110. So $\angle EDF = 110^\circ$? But option A is 70, D is 110. Wait, maybe I mixed up the angles.

Wait, another approach: $\triangle ABC$ is equilateral, so $\angle B = 60^\circ$. $EF \parallel AC$, so $\angle BEF = \angle BAC = 60^\circ$, $\angle BFE = \angle BCA = 60^\circ$, so $\triangle BEF$ is equilateral. Then $\angle DEF = \angle DFE$ (since $DE = DF$). $\angle DFC = 85^\circ$, so $\angle DFE = 180 - 85 = 95^\circ$? No, that can't be. Wait, no, $\angle BFE = 60^\circ$, so $\angle DFE = 180 - 60 - 85 = 35^\circ$? Wait, $B - F - C$, so $\angle BFE + \angle DFE + \angle DFC = 180^\circ$. $\angle BFE = 60^\circ$ (since $\triangle BEF$ is equilateral), $\angle DFC = 85^\circ$, so $\angle DFE = 180 - 60 - 85 = 35^\circ$. Then $\triangle DEF$ has $DE = DF$, so $\angle DEF = \angle DFE = 35^\circ$, so $\angle EDF = 180 - 35 - 35 = 110^\circ$. Yes, that makes sense. So the answer is D, 110°. Wait, but let's check again.

$\angle BFE = 60^\circ$ (equilateral triangle), $\angle DFC = 85^\circ$, so $\angle DFE = 180 - 60 - 85 = 35^\circ$. $\triangle DEF$ is isosceles with $DE = DF$, so $\angle DEF = \angle DFE = 35^\circ$. Then $\angle EDF = 180 - 35 - 35 = 110^\circ$. Yes, so the measure of $\angle EDF$ is $110^\circ$, which is option D. Wait, but earlier I thought $\angle FDC = 35^\circ$, but maybe that's not the case. Wait, no, the key is that $\angle BFE = 60^\circ$ (from equilateral $\triangle BEF$), so $\angle DFE = 180 - 60 - 85 = 35^\circ$, then $\angle EDF = 180 - 2 \times 35 = 110^\circ$. So the answer is 110°, option D. Wait, but let's confirm with the triangle sum.

In $\triangle DEF$, $DE = DF$, so it's isosceles with base $EF$. So base angles are $\angle DEF$ and $\angle DFE$. Sum of angles in triangle: $180^\circ$. So $\angle EDF + 2 \times \angle DFE = 180^\circ$. We need to find $\angle DFE$. Since $EF \parallel AC$, and $\triangle ABC$ is equilateral, $\angle BFE = 60^\circ$ (corresponding angles). Then, $\angle BFE + \angle DFE + \angle DFC = 180^\circ$ (linear pair on line $BC$). So $60^\circ + \angle DFE + 85^\circ = 180^\circ$. Solving for $\angle DFE$: $\angle DFE = 180 - 60 - 85 = 35^\circ$. Then $\angle EDF = 180 - 2 \times 35 = 110^\circ$. Yes, that's correct. So the answer is D, 110°.

Answer:

D. \( 110^\circ \)