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Two triangles ($ABC$ and $MNP$) are right (angles $ACB$ and $MPN$ are $90$ degrees each), the hypothenuses $AB = MN$ and the heights to the hypothenuses $CD = PQ$. Prove that the two triangles are congruent.

John
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3 Answers3

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Let $a$ and $b$ be the sides of ABC and $m$ and $n$ be the sides of MNP. Then, from the same hypotenuse and area, we have $$a^2+b^2=m^2+n^2$$ $$ab=mn$$

Two solutions due to symmetry: either $a=m,\>b=n$, or $a=n,\>b=m$, which yields congruent triangles either way.

Quanto
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You're given that the hypotenuses and the altitudes on them are congruent. Since the given segments determine the vertices of each triangle, they have determined both up to congruence, and the result follows.

Allawonder
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  • How are two of the sides immediately congruent? We have that one side of one triangle is equal to another side of another triangle. – John Oct 20 '19 at 13:41
  • @Gordon Oops, I misread the question a bit. It's one pair of congruent sides we're given, together with a pair of congruent altitudes -- I'd thought the latter were sides too. – Allawonder Oct 20 '19 at 13:43
  • No, we have a side and a height equal to a side an a height. – John Oct 20 '19 at 13:45
  • @Gordon Whether height or altitude, both are synonymous. – Allawonder Oct 20 '19 at 13:46
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    Can you give a more precise proof? This doesn't really follow. – John Oct 20 '19 at 13:47
  • @Gordon What do you mean by more precise? What's vague about my explanation? Finally, what doesn't follow about it? Did you bother to understand the proof at all? Or are you merely throwing about words you hardly know the meaning of? – Allawonder Oct 20 '19 at 13:48
  • I understand the proof, but I don't think "Since the given segments determine the vertices of each triangle" is precise in mathematical terms. – John Oct 20 '19 at 14:08
  • @Gordon If you know one side of a triangle (hypotenuse in this case), then you already know two of its vertices (the endpoints of the sides. If in addition you know the altitude to the given side, then you know the third vertex (the other endpoint of the altitude). Finally, if you know the three vertices of a triangle, then you know its sides, and hence the triangle. This is not more precise than before, just more verbose. – Allawonder Oct 20 '19 at 14:12
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Let $E$ be the midpoint of $AB.$ Let $R$ be the midpoint of $MN.$

Show that $\triangle CDE \cong \triangle PQR$, then show that either $\angle CAB \cong \angle PMN$ or $\angle CAB \cong \angle PNM.$

David K
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