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I need a hint on how to prove that "In a triangle ABC if D is the midpoint of the side AB and the line through D intersects AC at E such that BC=2DE then E is the midpoint of AC and BC is parallel to DE"

Thank you

rainthree
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  • Assume that $E$ is not the mid-point and prove by contradiction. – VVR May 28 '22 at 08:55
  • VVR, sure, I tried , but haven't succeeded. By the way the proof you posted and then deleted, has a mistake I think. In △ADE: ∢A+∢ADF+∢FDE=∢A+∢ADE=180°-∢AED not 180°-∢AFD as you wrote... And the proof hasn't used the info that DE=BC/2 so of course it is wrong, how could it be true.. – rainthree May 28 '22 at 09:31
  • I know that, that's the reason I deleted it – VVR May 28 '22 at 09:39
  • There is a problem: if AC > AB > BC then there's another solution with E not the midpoint of AC. Consider drawing an arc centred on D with radius ½BC; it will intersect AC at its midpoint and another point. – Martin Kealey May 28 '22 at 09:45
  • "There is a problem: if AC > AB > BC then there's another solution with E not the midpoint of AC. Consider drawing an arc centred on D with radius ½BC; it will intersect AC at its midpoint and another point." Can't visualize it, can you please show a drawing ? – rainthree May 28 '22 at 09:59
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    This has grown to a lot more than "a hint", so I'm not going to rush to draw something. Rather than "visualizing", try using a ruler, a compass, a pencil, and a sheet of paper. Hopefully you'll see it without me needing to draw it. – Martin Kealey May 31 '22 at 04:49

3 Answers3

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In $\triangle ABC$ , if we apply the law of sine, we get:

$$ \frac{BC}{\sin\angle BAC} = \frac{AB}{\sin\angle ACB}$$

or, $$\frac{BC}{AB} = \frac{\sin\angle BAC}{\sin\angle ACB} $$

Now, In $\triangle ADE$

$$ \frac{DE}{\sin\angle BAC} = \frac{AD}{\sin\angle AED}$$

or, $$ \frac{\frac{BC}{2}}{\sin\angle BAC} = \frac{\frac{AB}{2}}{\sin\angle AED}$$

or, $$\frac{BC}{AB} = \frac{\sin\angle BAC}{\sin\angle AED} $$

Therefore, $$\sin\angle ACB = \sin\angle AED$$

VVR
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  • Nice, thank you – rainthree May 28 '22 at 10:51
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    As I've alluded to elsewhere, sinφ = sinθ does not prove φ = θ, and indeed, the proof you seek does not exist for precisely this reason. But since my answer has been downvoted, I assume you're not interested in my helping you further. (Hint: you're only supposed to downvote answers that are wrong or misleading, not just because they're not "the best".) – Martin Kealey May 31 '22 at 03:29
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    @Martin Kealey, I agree with you, the question is not exactly the converse of mid-point theorem. My answer just proves that there are two possibilities, either $E$ is the mid-point of $AC$ or $\angle ACB + \angle AED = 180^\circ$, which is a completely different situation. – VVR May 31 '22 at 05:45
  • Martin, I haven't downvoted by the way. It must be other readers who did, and of course I am interested in you helping me further. If we add the condition that BC>AB, will this be enough to ensure that E will be the middle of AC ? – rainthree Jun 04 '22 at 09:46
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Edit:

a proof of the question as stated does not exist, as it's easy to construct a counterexample:

A circle of radius ½BC centred on D intersects (the infinite continuation of) AC at E and E’; for some triangles E’ will be between A and C, and therefore a line DE’ has length ½BC while not being parallel to BC.

E’ is a solution to "intersects at E" in the problem statement, which implies a "for all E" quantifier. If you're missing a "there exists" quantifier for E, then that quantifier would also allow satisfaction of "two-sides-and-any-angle makes a similar triangle", rebutting your objection in the comments below.

Original: Looks like a simple case of similar triangles, therefore the angles are the same, therefore...

Start by constructing DE as parallel, then show that it's half of BC. Then work backwards to form a complete proof.

Remember to confirm that parallellism is both necessary and sufficient (ie the interference is reversible).

Martin Kealey
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  • why are they similar ? we only know two pairs of proportional sides, we don't know that the pair of angles between these sides are congruent. – rainthree May 28 '22 at 08:51
  • You know a shared angle as well – Martin Kealey May 28 '22 at 08:56
  • "You know a shared angle as well" No, we don't. "Start by constructing DE as parallel, then show that it's half of BC." We are not allowed to assume that it is parallel. We only assume it is half of BC, and we have to prove that it is parallel – rainthree May 28 '22 at 08:57
  • The point is to run the proof in reverse, and then confirm that each step is reversible – Martin Kealey May 28 '22 at 08:58
  • The shared angle is ∡BAC = ∡DAF – Martin Kealey May 28 '22 at 09:00
  • That angle is not the angle formed by the sides which are proportional. ASS or SSA is not a similarity case – rainthree May 28 '22 at 09:02
  • The relevant proportionality is AD/AB = AE/AC. prove that BC∥DE if-and-only-if ABC and ADE are similar. – Martin Kealey May 28 '22 at 09:22
  • "prove that BC∥DE if-and-only-if ABC and ADE are similar." But we don't know they are similar. We don't know AD/AB=AE/AC . We only know AD/AB=DE/BC You claim that the other proof gets to this and it is reversible, please show the steps – rainthree May 28 '22 at 09:36
  • If they can be dissimilar then you have DISproved the original hypothesis (see comment on the question above). – Martin Kealey May 28 '22 at 09:47
  • Maybe they cannot be dissimilar, the question is how to prove they cannot be dissimilar. – rainthree May 28 '22 at 10:00
  • You can't, because a circle may intersect a line at TWO points. There are two solutions to "a line of length X from point P meeting line YZ" – Martin Kealey May 28 '22 at 11:18
  • If you want an interactive explanation, return to libera##math – Martin Kealey May 28 '22 at 11:20
  • "That angle is not the angle formed by the sides which are proportional. ASS or SSA is not a similarity case". There are at most two possible triangles given two sides and an angle; if there are two then one of them is similar and the other is not. These correspond to ADE and ADE’ in my revised explanation above. – Martin Kealey Jun 01 '22 at 00:45
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Name the midpoint of BC as F and join this point with the midpoint of AB (i.e. D). With this, use the midpoint theorem to conclude that DF is parallel to AC. With a little angle chasing and using one of the congruence theorems, we can prove that $\Delta ADE$ is congruent to $\Delta DBF$. From here, it will be obvious that DE is parallel to BC.

Swams2004
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  • I got that AD=DB, ∢DAE=∢BDF and DE=BF but it is not enough to use of the congruence postulates. How to chase one extra pair of congruent angles? – rainthree May 27 '22 at 17:26