Now construct a line from E perpendicular to AB. The point where it meets AB will be called F. Do the same on the other side: construct a line from E perpendicular to AC, and call the point of intersection G.

Now, ED is congruent to itself, and the angle EDB and EDC were both constructed as right angles, and DB is congruent to DC because D was chosen to bisect BC. Thus, the triangles EDB and EDC are congruent.

The angles EFA and EGA were both constructed as right angles, the angles FAE and GAE are the halves of a bisected angle, and of course AE is congruent to itself. Thus, the triangles FAE and GAE are congruent.

EB is congruent to EC, because they are corresponding parts of the congruent triangles EDB and EDC. Likewise, FE and GE are corresponding parts of FAE and GAE. Since the angles BFE and CGE are right angles, it follows that FEB and GEC are congruent (angle-side-side sufficing in the special case of a right angle).

Now, AF and AG are corresponding parts of FAE and GAE, and FB and GC are corresponding parts of FEB and GEC. Since AF=AG and FB=GC, AF+FB=AG+GC. Thus, AB=AC.

As a corolloary, it's simple to show that all triangles are equilateral, because it can be shown by the same method that AB=BC and BC=AC

Muahaha.. spend some time to figure out.. Draw it out or whatever.. There is no error with the proof...

---> Weirded weirdly by the weird weird Worksheet 3.142,

Labels: Worksheet 3.142