If two angles of a triangle are equal, prove that the triangle has a pair of equal sides. (construction - draw am, the bisector of anglebac, to meet bc at m)

Step-by-step explanation:

So we have angle B and C are equal and that AM is an angle bisector of BAC, that means BAM and MAC are both equal.  

Now This leaves us with two triangles BAM and CAM.  We know they all have the same angles.  ABM = ACM, BAM = MAC so since the third angle of each triangle  has to equal 180 minus the other two the third angle has to be equal  for both triangles.

If that was confusing, if we take ABM = ACM and  BAM = MAC and then we can say for the last angle:

AMC = 180 - ACM - MAC

AMC = 180 - ABM - BAM

AMC = AMB because AMB is  180 - ABM - BAM.  THough let me know if you don't get that.

Anyway, now we know AMC = AMB so we know it's a similar triangle.  Also you can always use this trick if there are ever two triangles where two angles are equal, you can show the third is equal as well, so if two are equal you know they're similar.

now we use properties of similar triangles.  When you have similar triangles, corresponding sides are scaled between them.  In this case the scale factor between these two is just 1, so all sides are equal.  Which means BA and AC are equal, and since they are two  sides of the original triangle the triangle has a pair of equal sides.  

Hopefully that helps, but let me know if i didn't explain anything well enough.  

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1