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Given isosceles $\triangle ABC$ with $ \angle A$ as the vertex angle. We know that the ff. are the same segment:

(1) the angle bisector of vertex angle $A$

(2) the median from vertex $A$

(3) the altitude from vertex $A$.

We also know that the converse of this proposition is again true.

A natural question arises: Given $\triangle ABC$. If a segment is two of the three segments mentioned above, is it also the third, i.e., is it enough for two of the three segments to be the same segment to conclude that $\triangle ABC$ is isosceles?

Assume (1) and (3) are the same segment. Then by ASA and CPCTC, the segment is also (2).

Next, assume (2) and (3) are the same segment. Then by SAS and CPCTC, the segment is also (1).

Lastly, assume (1) and (2) are the same segment. This is where the difficulty appears since ASS congruence is NOT valid. But I still think the claim is true.

Or are there counterexamples?

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If 1 and 2 are the same, then the angle bisect or theorem tells us that the third sides are equal, hence the triangle is isosceles.

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  • $\begingroup$ OF COURSE! Why didn't I think of that?! I guess was limited into using triangle congruence. Thanks Calvin! PS: I'm a big fan of brilliant.org. :D $\endgroup$ – Mark Lao Sep 28 '13 at 9:11

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