The triangle $PQR$ is equilateral. The circumference with centre $R$ and radius $r$ intersects the sides of the triangles at $S$ and $T$. Is the triangle $PQS$ congruent to the triangle $RQS$?


I'm having trouble with this. The solution says that they aren't congruent because they share only one side, but I can see they share sides $SQ$, and the sides $QR$ and $PQ$ are equal since the triangle is equilateral, because of this I think the solution might be wrong. Also, since the circumference intersects the triangle and I have the segment $QS$, then I think that segment should be a tangent of the circumference at $S$, making the angle $\angle{RSQ}=90$ degrees, thus making the angles $\angle{RSQ}$ and $\angle{QSP}$ equal, therefore making the triangles congruent ($SSA$), but I think my assumption might be wrong. Thanks.

  • $\begingroup$ Reflection at the height from $R$ to $PQ$ swaps $P\leftrightarrow Q$ and keeps the circle fixed, hence maps $PQS\leftrightarrow QPT$. -- But $RQS$ will in general not be congruent to $PQS$. This happens only if $r$ is half the side length $\endgroup$ – Hagen von Eitzen Jul 11 '17 at 14:54
  • $\begingroup$ SSA is not one of the congruence criteria for triangles. $\endgroup$ – DMcMor Jul 11 '17 at 14:55
  • $\begingroup$ @NickCassol Don't try to use ASS congruence... you'll make one of yourself. XD $\endgroup$ – Franklin Pezzuti Dyer Jul 11 '17 at 15:00
  • $\begingroup$ @DMcMor Why not? I'm translating the terms from Spanish, but I think it would still apply, two triangles are congruent if two sides are equal and the angle opposing the biggest of said sides are equal too, right? $\endgroup$ – Nick Cassol Jul 11 '17 at 15:41
  • $\begingroup$ @Nilknarf Why is that? Please explain. $\endgroup$ – Nick Cassol Jul 11 '17 at 15:42

They are not necessarily congruent, unless the radius is half of the triangle's side length. What if you drew the diagram like this?

enter image description here

... do they still look congruent?

  • $\begingroup$ No they don't, so they would be congruent only if the radius is half of the sides of the triangle, right? $\endgroup$ – Nick Cassol Jul 11 '17 at 15:43
  • $\begingroup$ @NickCassol Right. But that's not a given in your problem, so you can't assume that they are. $\endgroup$ – Franklin Pezzuti Dyer Jul 11 '17 at 15:43
  • $\begingroup$ Now I see, but was my assumption about the segment $QS$ being a tangent wrong?, I don't see why that'd be the case. $\endgroup$ – Nick Cassol Jul 11 '17 at 15:49

There are not congruent. Imagine like this,
Keep R and Q fixed. Moving along the line RP does not change the point S. So if my point P is anywhere along RP the triangle RQS is the same. But for various points P, we get different triangles PSQ.


They are congruent if $QS$ is tangent to the circle at $S$. For the infinitely many equilateral triangles with apex $R$ and $PR \ge SR$, $QS$ cuts the circle at two points, making $\angle RSQ$ and $\angle PSQ$ supplementary but not equal. Only when $QS$ is tangent at $S$ are the two angles equal and the triangles therefore congruent.

  • $\begingroup$ Because the segment $QS$ could intersect the circumference at two points, right? $\endgroup$ – Nick Cassol Jul 11 '17 at 16:15
  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – Sahiba Arora Jul 11 '17 at 16:24
  • $\begingroup$ @nick cassol--yes, and it will do that for all but one equilateral triangle, no? $\endgroup$ – Edward Porcella Jul 11 '17 at 17:14
  • $\begingroup$ @sahiba arora--OP seems to be asking whether his assumption or the solution is wrong. I was answering that his assumption is wrong. Should I explain why? $\endgroup$ – Edward Porcella Jul 11 '17 at 17:22
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    $\begingroup$ @sahiba arora--I follow your suggestion in the edit. $\endgroup$ – Edward Porcella Jul 11 '17 at 18:17

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