# The Final Pathetic Bleatings of the Forum

Question:

Andrej, I know it's trivial. This is why I am frustrated.
I'm in the 10th grade and it's Saturday and my parents won't
let me play online until I'm done with my homework so I'm
surfing researching for an english paper and realize I'm
"too busy surfing to solve my own problems" and came on
here. Can I please at least have a hint? If it weren't
trivial I would feel okay leaving it undone, but now I'm
grinding my teeth.

Replies:

You are two sick bunnies.

Ok, ok.

Give me a minute.

Suppose we are given a trapezoid ABCD where AB is the base, and suppose we know that the diagonals are congruent.

Given the vertices A and B, the height of the trapezoid, and the length of the diagonal AC, how many possibilities are there for the vertex C? At most two, call them C and C'.

Namely, C and C' are the intersections of a circle centered at A with radius equal to the length of the diagonal, and the straight line passing through C and D.

In the above picture, can C really be a vertex? No, because D would have to be to the left of C, but that would make the diagonal BD longer than the diagonal AC.

Which leads us to the conclusion that there is only one triangle ABC such that AC is the diagonal of a trapezoid. By a symmetric argument, we discover that there are two possible vertices D and D', and the one that is closer to A is the acceptable one.

The triangles ABC and ABD share the side AB, have the same altitude, and AC is congruent with BD. Give one side of a triangle, an altitude on that side, and another side, there are generally two solutions. But we have eliminated one of the solutions, hence the triangles ABC and ABD are congruent. Thus the trapezoid ABCD is isosceles.

Impossible.

I have a better solution.

I don't believe you.

Want to bet?

Sure, anytime.

If I win you give me a blow job. If you win, I give you a blow job.

So you admit that my geometric powers are greater than yours?

Never! The bet is on!

Ok.

Suppose a trapezoid ABCD with base AB has congruent diagonals. Let E be the intersection of the base AB with a line passing through D and parallel to the diagonal AC, as in the following picture:

Since EACD is a parallelogram and AC is congruent to BD by assumption, ED is congruent to BD. Therefore EBD is an isosceles triangle and so the angle AED is congruent with the angle ABD. At the same time, the angle BAC is congruent with AED because EACD is a parallelogram. Thus, the angles BAC and ABD are congruent. The triangles BAC and ABD have congruent adjacent sides and angles between those sides, hence they are congruent. Therefore, the side AD is congruent with the side BC, qed.

Hmm

Wait. I think I see a problem.

Hmmm.

Hmmm, no I think it's OK.

Just to be clear, my answer was not wrong qua wrong, but I admit

Yes?

Your solution possesses a certain elegance.

I AM THE MASTER OF MATHEMATICS!

I was afraid this might happen

SILENCE, MONKEY! WORSHIP ME AND MY SUPERIOR ABILITIES!

NOW... About that blow job. IT IS TIME FOR ME TO TAKE WHAT IS RIGHTFULLY MINE.

WRONG

What? You owe me some sweet, sweet loving baby.

I don't think so.

You see, there is one thing that you FAILED to consider.

Impossible.

As a citizen of Slovenia, I have DIPLOMATIC IMMUNITY! HAHAHAHAHAHAH

Curses.

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