If $XYZ$ is a right scalene triangle and if one of its constituent triangles is equilateral, then the other triangle must be isosceles. [closed]

This question is related to a component of a proof I constructed. I have posted the (pending) proof for verification on MSE: https://math.stackexchange.com/questions/2067151/use-the-information-given-to-construct-a-proof.

However, I would greatly appreciate it if someone could verify whether this specific statement I made is true:

If $XYZ$ is a right scalene triangle and if one of its constituent triangles is equilateral, then the other triangle must be isosceles (and vice-versa).

In the above proof, this statement was made to justify the following key question and abstract answer:

How can I show that a right scalene triangle consists of an equilateral triangle and an isosceles triangle?

Show that one of the triangles is an equilateral triangle.

In the above question, I also included the following diagram to illustrate the triangles:

Notice that the right scalene triangle consists of two other triangles when divided.

If anything is unclear, you may find clarification in the proof question posted above. Also, I will be glad to answer any questions if you'd prefer to ask me directly.

If my reasoning is incorrect, please explain why.

Thank you.

closed as unclear what you're asking by Jean Marie, Rohan, Henrik, E. Joseph, John BDec 23 '16 at 9:25

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• What is $W$ here\? – Qwerty Dec 23 '16 at 6:51
• @Qwerty $W$ is the midpoint of the hypotenuse. – The Pointer Dec 23 '16 at 6:52
• What does bring the adjective "scalene" in the title ? Do you mean by this "any right triangle" ? – Jean Marie Dec 23 '16 at 7:01
• @ThePointer In any right triangle the midpoint of the hypothenuse is the center of the circumscribed circle, so $WX = WY \implies WX = WY=WZ$. This was addressed in the original question and answer. – dxiv Dec 23 '16 at 7:06
• You didn't really include any reasoning in this posted question. if one of its constituent triangles is equilateral, then the other triangle must be isosceles (and vice-versa) The vice-versa part is obviously false, take an isosceles right triangle for example. – dxiv Dec 23 '16 at 7:12