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I'm having some trouble with the following problem:

Prove that a triangle that has two congruent angles is isosceles

I tried to prove this by separating it into two triangles and use the ASA or the SAS postulate. However, I am stuck. I need some help. Thank you!

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  • $\begingroup$ A fancy way to do it is the sine rule. $\endgroup$ – Sawarnik Mar 6 '14 at 6:21
  • $\begingroup$ Isn't that the way we define an isosceles triangle? $\endgroup$ – AJMansfield Mar 6 '14 at 15:15
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We can do it without drawing any line.

Let our triangle be $ABC$, with $\angle B=\angle C$.

By ASA, $\triangle ABC$ and $\triangle ACB$ are congruent.

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  • $\begingroup$ Wow, that's a simple proof! $\endgroup$ – user122283 Mar 6 '14 at 1:33
  • $\begingroup$ This is very nice proof. Thank you $\endgroup$ – eChung00 Mar 6 '14 at 1:43
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    $\begingroup$ I recall that many decades ago, a few researchers were creating an artificial intelligence program that would prove statements in geometry. The program came up with this proof, which was unknown to the researchers. $\endgroup$ – Joel Reyes Noche Mar 6 '14 at 1:44
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    $\begingroup$ Very interesting, @JoelReyesNoche... you wouldn't happen to have any links or citations for that research? I would love to read more on that. $\endgroup$ – buruzaemon Mar 6 '14 at 2:20
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    $\begingroup$ I guess it is a question of notation. When I use the terms congruent, or similar, and list the vertices, it is in corresponding order. $\endgroup$ – André Nicolas Mar 6 '14 at 11:50
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Hint: Bisect the angle that is different. The two halves are congruent.

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Edit: My original answer was incorrect. Here is the correct version as given by Marc van Leeuwen in a comment:

There is no need to bisect angles. Let triangle $\triangle ABC$ have $\angle A=\angle B$. Since $\angle A=\angle B$, $\overline{AB}=\overline{BA}$, and $\angle B=\angle A$, then $\triangle CAB$ is similar to $\triangle CBA$. Thus, $\overline{BC}=\overline{AC}$.

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    $\begingroup$ Not really. While André Nicholas is vague about which angles and sides are compared when applying ASA (one easily guesses), you actually point at the wrong things. It is not the top angle that is compared to itself, nor one of the equal sides; you need to use the hypothesis of equal angles. It should be: since $\angle A=\angle B$ and $\overline{AB}=\overline{BA}$ and $\angle B=\angle A$,... $\endgroup$ – Marc van Leeuwen Mar 6 '14 at 5:54
  • $\begingroup$ @MarcvanLeeuwen, you are correct. Thank you very much for this. I am editing my answer to include your comment. $\endgroup$ – Joel Reyes Noche Mar 6 '14 at 7:08
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Hint: Draw a perpendicular bisector through the third side, the one that is not congruent to another, then show the two triangles formed are congruent.

Use ASA using the fact that a perpendicular bisector is a median. It then follows that the two other sides are congruent because they are corresponding parts of congruent triangles.

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  • $\begingroup$ How do we know that we can draw perp bisector from the vertex? I think there might be a case when a line bisect the angle but cannot be perp to the base. $\endgroup$ – eChung00 Mar 6 '14 at 1:32
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    $\begingroup$ @eChung00 In a acute angled triangle, it is pretty obvious to show that a line from the vertex to the other side is within the triangle. The perpendicular bisector part follows. $\endgroup$ – user122283 Mar 6 '14 at 1:34
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If you bisect the vertex angle, you find that you have created two congruent triangles. The triangles are congruent because of AAS congruence. Because of CPCTC, the sides are congruent as well. It is hard to describe. See this image:

enter image description here

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  • $\begingroup$ Is AAS triangles are congruent? I only know ASA, SSS, and SAS. $\endgroup$ – eChung00 Mar 6 '14 at 1:33
  • $\begingroup$ Yes. You cannot make the proof without knowledge of that congruence postulate. As you can see, the two angles are angle B congruent to angle C (given) and angle BMA congruent to CMA because we have constructed the perpedicular bisector. Also, AM is congruent to AM, giving us 3 consecutive angles. $\endgroup$ – louie mcconnell Mar 6 '14 at 1:36
  • $\begingroup$ Doesn't matter. 1st, I created a perpendicular bisector, so it would create a right angle. Second, even if you state that BM isn't congruent to CM, it still works, because angle BAM is congruent to CAM because we bisected them, angle B is congruent to C, and AM is congruent to AM. $\endgroup$ – louie mcconnell Mar 6 '14 at 1:39
  • $\begingroup$ @eChung00 - since a triangle only has 3 angles, AAS is in effect, saying ASA. Right? If two angles are congruent, the third must me. I suppose louie should state that for an elegant proof, but it's correct. $\endgroup$ – JoeTaxpayer Mar 6 '14 at 1:39
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Say the triangle is $\Delta ABC$. Let the congruent angles be $\angle B,\angle C$. Then, we have to show that $\overline{AB}\cong\overline{AC}$. To do this, connect the angle bisector of $\angle A$ with point $A$ to get $\overline{DA}$ ($D$ is where the angle bisector intersects $\overline{BC}$). Then, $\overline{DA}\cong \overline{DA}$, and $\angle B\cong\angle C$. So, $\angle BAD=\angle CAD$. By the AAS postulate, $\Delta ACD\cong\Delta ABD$. Since corresponding sides of congruent triangles are congruent, $\overline{AC}\cong\overline{AB}$. Q.E.D.

I apologize for not being able to draw the diagram.

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  • $\begingroup$ I am confused why is this ASA. The S part should not be DA I think. $\endgroup$ – eChung00 Mar 6 '14 at 1:37
  • $\begingroup$ @eChung00 You have $\overline{DA}\cong\overline{DA}$, $\angle B\cong\angle C,\angle BAD=\angle CAD$. This forms AAS, on drawing. $\endgroup$ – user122283 Mar 6 '14 at 1:41
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The equality of two sides opposite to two congruent angles in a triangle is obvious from the law of sines.

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