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When I learned this fact, everyone just took it for granted, and no one attempted to prove it. I've seen the proof using the law of sines, but is there a way of proving this without trigonometry?

pavle
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  • A quick google for "proof AAA similarity triangle" gave this. You are looking for a proof like that? – M. Winter Jul 06 '18 at 22:10
  • Not really, I looked at the proof, and it was fine until it mentioned the basic proportionality theorem, and on that page there is no proof of it, so I'm basically looking for the proof of the basic proportionality theorem in that case. @M.Winter – pavle Jul 06 '18 at 22:20
  • The whole premise is based on angle magnitude being a ratio of the triangle's sides irrespective of the actual lengths. If you change that ratio, you change the angle. – Phil H Jul 06 '18 at 22:34
  • Are you willing to accept that opposite sides of a parallelogram are congruent? – Somos Jul 07 '18 at 01:33
  • @AyyLmao A proof of the basic proportionality theorem can be found on the same site: https://www.ask-math.com/basic-proportionality-theorem.html That ultimately relies on the area formula for triangle (base times height by 2) – Hagen von Eitzen Jul 07 '18 at 06:51

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Here's a proof of the basic proportionality theorem that relies "only" on congruences and betweenness.

Let $ABC$ and $A'B'C'$ be triangles with congruent angles. By moving the second triangle, we may assume wlog that $A'=A$ and $B'$ is on the ray $AB$ and that $C'$ is in the same half plane determined by $AB$ as $C$. Then from $\angle CBA=\angle C'B'A'$, we see that $BC\|B'C'$. And from $\angle BAC=\angle B'A'C'$, we see that $C'$ is on the ray $AC$. Wlog. $B'$ is between $A$ and $B$. Then also $C'$ is between $A$ and $C$.

Pick a (large) integer $n$ and partition $AB$ into $n$ equal parts by points $P_0=A,P_1,\ldots, P_n=B$. Likewise, partition $AC$ into $n$ equal parts by points $Q_0=A,Q_1,\ldots, Q_n=C$. The lines trough the $P_i$ parallel to $AC$ and the lines through the $Q_i$ parallel to $AB$ produce an $n\times n$ grid of parallograms that are readily seen to be congruent. In particular, the lines $P_iQ_i$ ($i>0$) form diagonals of many of those parallelograms, splitting them into pairwise congruent tiny triangles. Simply by counting how many edges of tiny triangles contribute to the line segments, we find $$|P_0P_i|:|P_0P_j|= |Q_0Q_i|:|Q_0Q_j|=|P_iQ_i|:|P_jQ_j|.$$ If $B'$ is between $P_k$ and $P_{k+1}$, we see from $B'C'\|P_kQ_k$ that $C'$ is between $Q_k$ and $Q_{k+1}$. Then from parallelograms with two sides given by $B'C'$ and either $P_kB'$ or $B'P_{k+1}$, one finds $$|P_kQ_k|\le |B'C'|\le|P_{k+1}Q_{k+1}|. $$ Now we have $$k:n\le |A'B'|:|AB|\le (k+1):n $$ $$k:n\le |A'C'|:|AC|\le (k+1):n $$ $$k:n\le |B'C'|:|BC|\le (k+1):n $$ As $n$ was arbitrary, we conclude $$|A'B'|:|AB|=|A'C'|:|AC|=|B'C'|:|BC|. $$

Hagen von Eitzen
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