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Let’s solve each proof step by step. We’ll use the given information and geometry rules to prove what’s asked.

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Proof #1:

Given:
- E is the midpoint of AC → So AE = EC
- E is the midpoint of BD → So BE = ED

We need to prove: △AEB ≅ △CED

Step-by-step reasoning:

1. Since E is the midpoint of AC, then AE ≅ EC (by definition of midpoint).
2. Since E is the midpoint of BD, then BE ≅ ED (also by definition of midpoint).
3. Look at angles ∠AEB and ∠CED — they are vertical angles (formed where two lines cross). Vertical angles are always congruent → ∠AEB ≅ ∠CED.
4. Now we have:
- Side AE ≅ side EC
- Angle ∠AEB angle ∠CED
- Side BE ≅ side ED
5. That’s Side-Angle-Side (SAS) congruence! Two sides and the included angle match.

✔ So, △AEB ≅ △CED by SAS.

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Proof #2:

Given:
- I is on JH̄
- K is the midpoint of IJ̄ → so IK = KJ
- L is the midpoint of IH̄ → so IL = LH

We need to prove: JK̄ ≅ HL̄

Step-by-step reasoning:

1. K is midpoint of IJ̄ → IK ≅ KJ → so JK = ½ IJ
2. L is midpoint of IH̄ → IL ≅ LH → so HL = ½ IH
3. But wait — we don’t know if IJ = IH. However, look again: The diagram shows points J-I-H in a line? Actually, from the figure, it seems J, I, H are colinear with I between J and H? Not necessarily stated.

Wait — let’s reread: “I is on JH̄” → that means point I lies somewhere on segment JH. So J—I—H are in a straight line, with I between J and H.

So JH is one big segment, and I is somewhere on it.

Then:

- K is midpoint of IJ → so JK = KI = ½ IJ
- L is midpoint of IH → so IL = LH = ½ IH

But we want to prove JK ≅ HL.

That would only be true if IJ = IH — which isn’t given.

Wait — maybe there’s a typo or misread? Let me check the original problem again.

Actually, looking back: In Proof #2, the goal is to prove JK̄ ≅ HL̄

But unless IJ = IH, this won’t hold. Unless... perhaps the diagram implies symmetry? Or maybe we’re supposed to assume something else?

Wait — no, actually, let’s think differently.

Maybe the key is that both JK and HL are halves of segments that together make up JH? But still, without knowing how I divides JH, we can’t say JK = HL.

Hold on — perhaps I misread the givens.

Original says:

> Given: I is on JH̄, K is the midpoint of IJ̄, L is the midpoint of IH̄
> Prove: JK̄ ≅ HL̄

This is only true if IJ = IH — i.e., if I is the midpoint