# Are these triangles congruent?

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.

• 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 – Hagen von Eitzen Jul 11 '17 at 14:54
• SSA is not one of the congruence criteria for triangles. – DMcMor Jul 11 '17 at 14:55
• @NickCassol Don't try to use ASS congruence... you'll make one of yourself. XD – Franklin Pezzuti Dyer Jul 11 '17 at 15:00
• @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? – Nick Cassol Jul 11 '17 at 15:41
• @Nilknarf Why is that? Please explain. – 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?

... do they still look congruent?

• No they don't, so they would be congruent only if the radius is half of the sides of the triangle, right? – Nick Cassol Jul 11 '17 at 15:43
• @NickCassol Right. But that's not a given in your problem, so you can't assume that they are. – Franklin Pezzuti Dyer Jul 11 '17 at 15:43
• 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. – 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.

• Because the segment $QS$ could intersect the circumference at two points, right? – Nick Cassol Jul 11 '17 at 16:15
• 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 – Sahiba Arora Jul 11 '17 at 16:24
• @nick cassol--yes, and it will do that for all but one equilateral triangle, no? – Edward Porcella Jul 11 '17 at 17:14
• @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? – Edward Porcella Jul 11 '17 at 17:22
• @sahiba arora--I follow your suggestion in the edit. – Edward Porcella Jul 11 '17 at 18:17