Question 970823
Show by means of angles that A=(-1,0), B=(4,6), and C=(10,1) are the vertices of an isosceles triangle.
<pre>
We draw the triangle to see which angles appear to be the base angles.

{{{drawing(400,325,-3,13,-3,10, graph(400,325,-3,13,-3,10),
circle(-1,0,0.15),circle(-1,0,0.13),circle(-1,0,0.11),circle(-1,0,0.09),circle(-1,0,0.07),circle(-1,0,0.05),circle(-1,0,0.03),circle(-1,0,0.01),
triangle(-1,0,4,6,10,1),locate(-3,.8,"A(-1,0)"),locate(10.2,1.5,"C(10,1)"),
circle(4,6,0.15),circle(4,6,0.13),circle(4,6,0.11),circle(4,6,0.09),circle(4,6,0.07),circle(4,6,0.05),circle(4,6,0.03),circle(4,6,0.01),locate(4.2,6.4,"B(4,6)"),

circle(10,1,0.15),circle(10,1,0.13),circle(10,1,0.11),circle(10,1,0.09),circle(10,1,0.07),circle(10,1,0.05),circle(10,1,0.03),circle(10,1,0.01) )}}}


We would normally use distances to show that AB &#8773; AC to show that ABC is
isosceles. But we are instructed not to do it that way, but to show 
instead that &#8736;A &#8773; &#8736;C.

First we find the slopes of all three sides:

Slope of AB:  {{{m[AB]=(6-0)/(4-(-1))=6/(4+1)=6/5}}}  
Slope of AC:  {{{m[AC]=(1-0)/(10-(-1))=1/(10+1)=1/11}}}
Slope of BC:  {{{m[BC]=(1-6)/(10-4)=(-5)/6=-5/6}}}

The base angles are acute angles so we use absolute value
to insure that the tangent is positive:

{{{tan(A)=abs((m[AB]-m[AC])/(1+m[AB]m[AC]))=abs((6/5-1/11)/(1+expr(6/5)*(1/11)))}}}

Multiply top and bottom by 55

{{{tan(A)=abs((66-5)/(55+6))=61/71}}}

Now we do that for &#8736;C:

{{{tan(C)=abs((m[AC]-m[BC])/(1+m[AC]m[BC]))=abs((1/11-(-5/6))/(1+expr(1/11)*(-5/6)))=abs((1/11+5/6)/(1-5/66))}}}

Multiply top and bottom by 66

{{{tan(C)=abs((6+55)/(66+5))=61/71}}}

So &#8736;A and &#8736;C have the same tangents and so they have the same measure
and since the base angles of &#916;ABC are &#8773;, &#916;ABC is isosceles.

Edwin</pre>