In the lecture of geometry, we had discussed various problems on proving properties of an isosceles triangle for a long time. In the last lecture we had its final argument.

(Today I was so tired of drawing pictures with a vector tool that I drew them by moving mouse on a bitmap tool).

The property to prove was the following:

*If the two of three angles of a triangle are equal in measure, the triangle is an isosceles.*

At first, we drew a figure:

With the symbols above, we write the axiom and conclusion:

**axiom**

In the above triangle ABC, angle CAB = angle CBA

**conclusion**

CA=CB

[Failed Case 1] To set the midpoint M on the side AB and join M and C

Denote M on the side AB to be AM=BM, that is, M is the midpoint of AB.

Join C and M, then the triangle ABC is divided into two triangles, CAM and CBM.

If CAM and CBM are congruent, then CA would be equal to CB. Is it true?

AM=BM by the definition of M.

CM is identical.

But we can’t prove easily that the angle CMA would be equal to CMB.

So the proof in this way might be complicated.

[Failed Case 2] To draw the ray from C perpendicular to AB

If you draw the ray from C perpendicular to the side AB….

No, it is not yet proved that you can really do that, because there might be a case that ray goes to the extension of the segment AB.

[The Forgiven Case] To draw the bisector of the angle ACB

The professor told finally, it is permitted to use without proof, to draw the bisector of the angle ACB and get its intersection on the side AB.

Well, it was for me understandable. If the ray from C goes **outside** of the triangle, this ray cannot go **inside** the angle to divide it. That might not require a proof…

Therefore, denoting D as the intersection of the bisector of the angle ACB with the side AB, we could say,

AC=AC(identical)

angle ACD = angle BCD (the definition)

angle ADC = 180 degrees – angle ACD – angle CAD

= 180 degrees – angle BCD – angle CBA =angle BDC

Then we use the ASA rule to prove that the triangle ACD is congruent to the triangle BCD

So AC= BC

This problem was so difficult to me, because I had not been able to decide, which property was required a proof and which was not.

Therefore our discussion on an isosceles triangle ended.

Next time we will learn about a parallelogram.