Question 690163
<font face="Times New Roman" size="+2">


Good start.  Triangle APC is a right triangle.  Using similar logic, show that Triangle APB is also right.  Then since AC congruent to AB and AP congruent to AP by reflexive equality, the two right triangles are congruent by the Hypotenuse-Leg Postulate.  Then PB is congruent to PC by CPCT and finally triangle BPC is isosceles by definition.  What about the case where BC is also congruent to BP and CP?  No problem, the set of isosceles triangles does not exclude equilateral triangles.  By the way, you can get rid of that bit about angles ABC and ACB being congruent.  They are, but nobody cares and it doesn't help you get where you are going.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>