Question

triangle efg is formed by connecting the midpoints of the side of triangle bcd. the measures of the interior angles of triangle bcd are shown. find the measure of ∠dfg. figures not necessarily drawn to scale.

Explanation:

Step1: Find angle at C in triangle BCD

In triangle \(BCD\), the sum of interior angles is \(180^\circ\). Given \(\angle B = 44^\circ\) and \(\angle D = 53^\circ\), so \(\angle BCD=180^\circ - 44^\circ - 53^\circ = 83^\circ\) (wait, no, wait the diagram has \(\angle C = 83^\circ\)? Wait, maybe first find angle at \(B\), \(C\), \(D\) sum. Wait, the problem says triangle \(BCD\) has angles: \(\angle B = 44^\circ\), \(\angle C = 83^\circ\), \(\angle D = 53^\circ\)? Wait no, let's recalculate. Sum of angles in triangle is \(180^\circ\). So \(\angle B + \angle C + \angle D = 180^\circ\). If \(\angle B = 44^\circ\), \(\angle D = 53^\circ\), then \(\angle C = 180 - 44 - 53 = 83^\circ\), which matches the diagram's \(\angle C = 83^\circ\).

Step2: Determine midsegments (Midline Theorem)

Since \(E\), \(F\), \(G\) are midpoints (triangle \(EFG\) is formed by connecting midpoints of \(BCD\)), so \(FG\) is a midline of triangle \(BCD\), meaning \(FG \parallel BC\) (by Midline Theorem: the segment connecting midpoints of two sides is parallel to the third side and half its length).

Step3: Find \(\angle DFG\) (corresponding angles)

Since \(FG \parallel BC\), \(\angle DFG\) and \(\angle BCD\) are corresponding angles (because \(FD\) is a transversal). Wait, no: \(FG \parallel BC\), so \(\angle DFG\) corresponds to \(\angle C\)? Wait, let's see the sides. \(F\) is midpoint of \(CD\)? Wait, no, the problem says "connecting the midpoints of the side of triangle \(BCD\)". So \(E\) (midpoint of \(BC\)), \(F\) (midpoint of \(CD\)), \(G\) (midpoint of \(BD\))? Then \(FG\) connects midpoints of \(CD\) and \(BD\), so \(FG \parallel BC\) (Midline Theorem: midline parallel to \(BC\) and half its length). Therefore, \(\angle DFG\) and \(\angle BCD\) are corresponding angles (since \(FG \parallel BC\) and \(CD\) is transversal), so \(\angle DFG = \angle BCD\). Wait, but \(\angle BCD\) is \(83^\circ\)? Wait no, wait in triangle \(BCD\), \(\angle B = 44^\circ\), \(\angle D = 53^\circ\), so \(\angle BCD = 180 - 44 - 53 = 83^\circ\), which matches the diagram's \(\angle C = 83^\circ\). So \(\angle DFG = \angle BCD = 83^\circ\)? Wait, no, that can't be. Wait, maybe I mixed up the sides. Wait, \(F\) is midpoint of \(CD\), \(G\) is midpoint of \(BD\), so \(FG \parallel BC\), so \(\angle DFG\) is equal to \(\angle C\) (corresponding angles). So \(\angle DFG = 83^\circ\)? Wait, but let's check again. Wait, maybe the angle at \(B\) is \(44^\circ\), angle at \(D\) is \(53^\circ\), angle at \(C\) is \(83^\circ\). Then \(FG \parallel BC\), so \(\angle DFG = \angle C = 83^\circ\)? Wait, no, that seems off. Wait, maybe I made a mistake. Wait, the Midline Theorem: in triangle \(BCD\), if \(F\) is midpoint of \(CD\) and \(G\) is midpoint of \(BD\), then \(FG \parallel BC\) and \(FG = \frac{1}{2}BC\). Therefore, \(\angle DFG\) and \(\angle BCD\) are corresponding angles, so they are equal. So \(\angle DFG = \angle BCD = 83^\circ\)? Wait, but let's recalculate the angle at \(C\) again. Sum of angles in triangle: \(44 + 53 + 83 = 180\), yes, \(44+53=97\), \(97+83=180\). So that's correct. So \(\angle DFG = 83^\circ\)? Wait, but maybe the question is different. Wait, no, maybe \(FG\) is parallel to \(BC\), so \(\angle DFG = \angle C = 83^\circ\). Alternatively, maybe I messed up the midpoints. Wait, the problem says "triangle \(EFG\) is formed by connecting the midpoints of the side of triangle \(BCD\)". So midpoints of \(BC\), \(CD\), \(BD\) – so \(E\) (mid \(BC\)), \(F\) (mid \(CD\)), \(G\) (mid \(BD\)). Then \(FG\) is midline of \(BCD\), parallel to \(BC\), so \(\angle DFG = \angle BCD = 83^\circ\). Wait, but let's check the angles again. Wait, maybe the angle at \(B\) is \(44\), angle at \(D\) is \(53\), so angle at \(C\) is \(83\), so yes, \(\angle DFG = 83^\circ\). Wait, but maybe the question is to find \(\angle DFG\) as equal to \(\angle B\)? No, that doesn't make sense. Wait, maybe I made a mistake in the Midline Theorem. Wait, Midline Theorem: the segment connecting midpoints of two sides is parallel to the third side. So if \(F\) is midpoint of \(CD\) and \(G\) is midpoint of \(BD\), then \(FG \parallel BC\), so \(\angle DFG\) and \(\angle BCD\) are corresponding angles, so they are congruent. Therefore, \(\angle DFG = \angle BCD = 83^\circ\). Wait, but let's confirm with the sum. Alternatively, maybe the triangle \(EFG\) is similar to \(BCD\), but since it's midpoints, it's a midline triangle, so similar with ratio \(1/2\), so angles are equal. So \(\angle DFG = \angle C = 83^\circ\). Wait, but the diagram shows \(\angle C = 83^\circ\), so that matches. So the measure of \(\angle DFG\) is \(83^\circ\)? Wait, no, wait, maybe I mixed up the angles. Wait, \(\angle B = 44\), \(\angle D = 53\), so \(\angle C = 180 - 44 - 53 = 83\), correct. Then \(FG \parallel BC\), so \(\angle DFG = \angle C = 83^\circ\). Yes, that makes sense.

Answer:

\(83^\circ\)