Proofs without Words 67

“Proof” of proposition 47. Consider an arbitrary triangle ABC.

P

R

Q

S

B

A

C

Let Q be the intersection of the angle bisector (blue) at ∠A and the perpendicular bisector

(green) of BC at midpoint P. Drop perpendiculars from Q to AB at R and to AC at S .

Because P is the midpoint of BC and PQ is perpendicular, we deduce that BQ  CQ by

the Pythagorean theorem. Since AQ is the angle bisector of ∠A, the right triangles AQR

and AQS are similar, and since they share a hypotenuse, they are congruent. It follows that

AR  AS and also QR  QS . Therefore, BQR is congruent to CQS by the hypotenuse-

leg congruence theorem. So RB  SC. And therefore,

AB  AR + RB  AS + SC  AC,

and so the triangle is isosceles, as desired.

I once presented the proof argument above to an advanced mathematics research seminar

at the University of California at Berkeley, as entertainment while we were waiting for the

scheduled speaker. When the speaker arrived, however, a senior member of the audience

insisted that we first get to the bottom of the puzzle, which had sent him into a fit of

confusion. Are you confused? Why not?