Given that:

ABC is an isosceles triangle in which

    AB = AC

The lengths of the following segmets are equal:

    AD
    DE
    EC
    BC

Find the measure of angle A.

I tried approaching the problem from a geometrical point of view, instead of a trigonometrical one. I started with the principles that the interior angles of a triangle equal a straight angle (180°), and the base angle of an isosceles triangle are equal.

There was no division by any terms with variables, but still, the only answer I found was 60°, which is the degenerate case.

I now suspect that despite appearances, the arrangement is impossible, and that is what we need to show. I could probably even prove it to be impossible using calculus if I tried.

But Vohanam's comments still suggest that there is an AHA! approach I'm missing, so I'll keep working on it.

Posted by TomM on 2002-06-18 16:51:28