Question

which additional piece of information would allow you to prove that the triangles are congruent by the hl theorem? a ( mangle dfe = 40 ) c ( overline{ab} cong overline{de} ) b ( mangle f = mangle abc ) d ( overline{ac} cong overline{df} ) a d c b

Explanation:

To prove congruence by the HL (Hypotenuse - Leg) theorem for right - angled triangles, we need a right angle, a pair of congruent hypotenuses, and a pair of congruent legs.

1. First, identify the right - angled triangles:

• Triangles \(ABC\) and \(DEF\) are right - angled triangles with \(\angle B=\angle E = 90^{\circ}\), and we can see from the diagram that \(AC\) and \(DF\) are the hypotenuses (the sides opposite the right angles) and \(BC\) and \(EF\) are marked as congruent (the legs).

1. Analyze the HL theorem requirements:

• The HL theorem states that if the hypotenuse and one leg of a right - triangle are congruent to the hypotenuse and one leg of another right - triangle, then the two triangles are congruent.
• We already know that one pair of legs (\(BC\cong EF\)) and the right angles (\(\angle B=\angle E = 90^{\circ}\)) are present. For HL, we need the hypotenuses to be congruent or another pair of legs and the hypotenuses. But looking at the options:
• Option A: \(m\angle DFE = 40\) gives an angle measure, which is not related to the HL theorem (HL is about sides, not angles for the congruence condition here).
• Option B: \(m\angle F=m\angle ABC\) gives angle - angle information, which is for AAS or ASA, not HL.
• Option C: \(\overline{AB}\cong\overline{DE}\): \(AB\) and \(DE\) are legs, but we already have one pair of legs (\(BC\cong EF\)) marked. For HL, we need the hypotenuse and a leg. If we take \(AB\cong DE\) and \(BC\cong EF\), that would be SAS, not HL.
• Option D: \(\overline{AC}\cong\overline{DF}\): \(AC\) and \(DF\) are the hypotenuses of the right - angled triangles \(ABC\) and \(DEF\) respectively. Since we already have \(BC\cong EF\) (legs) and \(\angle B=\angle E = 90^{\circ}\) (right angles), when we have \(AC\cong DF\) (hypotenuses) and \(BC\cong EF\) (legs), we can apply the HL theorem. Also, if we consider the other way, if we have the hypotenuse \(AC\cong DF\) and the leg \(BC\cong EF\) (given by the marking in the diagram), then by HL, \(\triangle ABC\cong\triangle DEF\).

Answer:

C. \(\overline{AB}\cong\overline{DE}\) (Wait, no, correction: The correct option is D. \(\overline{AC}\cong\overline{DF}\). Wait, let's re - evaluate.

Wait, the right - angled triangles: \(\triangle ABC\) with right angle at \(B\), hypotenuse \(AC\); \(\triangle DEF\) with right angle at \(E\), hypotenuse \(DF\). We can see that \(BC\) and \(EF\) are marked as equal (the legs). For HL, we need hypotenuse and a leg. So if \(AC\cong DF\) (hypotenuses) and \(BC\cong EF\) (legs), then by HL, the triangles are congruent.

So the correct option is D. \(\overline{AC}\cong\overline{DF}\)

So the answer is D. \(\overline{AC}\cong\overline{DF}\) (the option D, so in the multiple - choice context, the answer is the option with \(\overline{AC}\cong\overline{DF}\), which is option D. So the answer is D. \(\overline{AC}\cong\overline{DF}\) (the option D in the given choices, so the answer is the option labeled D, i.e., D. \(\overline{AC}\cong\overline{DF}\))