Question 1200932
<pre>
{{{drawing(400,350,-.5,2.5,-.5,2.5,
locate(0,0,A), locate(2,0,B), locate(1.51,.97,D), locate(1,1.9,C),
triangle(0,0,2,0,1,sqrt(3)), triangle(0,0,2,0,3/2,sqrt(3)/2) )}}}

Given: 
In △ ABC, segment AC = segment BC and segment AB ≠ segment AC. segment AD is a median.

Prove: segment AD cannot be perpendicular to segment BC

Assume (for contradiction) that AD is perpendicular to BC

CD = BD                                  because AD is a median
AD is perpendicular to BC                by assumption (for contradiction)
Angle ADC = Angle ABD                    Because AD perp. to BC
AD = AD                                  symmetric property 
triangle ACD congruent to triangle ABD   SAS
AB = AC                                  C.P.C.T.
Contradiction to segment AB ≠ segment AC. 


Edwin</pre>