Why do triangles with congruent angles have the same corresponding side ratios?

Question

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?

Answer 1

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 $$|A'B'|:|AB|=|A'C'|:|AC|=|B'C'|:|BC|. $$