Question 814754
<pre>
Let's plot it and see:

{{{drawing(400,4400/13,-2,11,-2,9,graph(400,4400/13,-2,11,-2,9),
locate(2,3,"D(2,3)"),locate(5,7.6,"E(5,7)"), locate(9,4,"F(9,4)"),
triangle(2,3,5,7,9,4) )}}}

It's pretty clear that angles D and F are acute.  Hoever angle E
looks suspiciously like it just might be a right angle.  To find out
we need to find the slopes of DE and EF to find out if they are 
perpendicular.  To find that out we find whether the slope of 
one is the reciprocal of the other with the sign changed.

Find the slope of DE

m = {{{(y[2]-y[1])/(x[2]-x[1])}}}

where (x<sub>1</sub>,y<sub>1</sub>) = (2,3)
and where (x<sub>2</sub>,y<sub>2</sub>) = (5,7)

m = {{{(7-3)/(5-2)}}} = {{{4/3}}}.

Find the slope of EF

m = {{{(y[2]-y[1])/(x[2]-x[1])}}}

where (x<sub>1</sub>,y<sub>1</sub>) = (5,7)
and where (x<sub>2</sub>,y<sub>2</sub>) = (9,4)

m = {{{(4-7)/(9-5)}}} = {{{(-3)/4}}} = {{{-3/4}}}

So oe slope is {{{4/3}}} and the other is {{{-3/4}}},
which is the recipocal of {{{4/3}}} with the sign 
changed.

So the  sides DE and EF are perpendicular and therefore
the triangle is a right triangle.

It also looks like it might be isosceles, with DE and EF
equal. so let's find the lengths of the legs DE and EF 

d = &#8730;<span style="text-decoration: overline">(x<sub>2</sub>-x<sub>1</sub>)²+(y<sub>2</sub>-y<sub>1</sub>)²</span>

For DE

d = &#8730;<span style="text-decoration: overline">(5-2)²+(7-3)²</span>  

d = &#8730;<span style="text-decoration: overline">(3)²+(4)²</span>

d = &#8730;<span style="text-decoration: overline">9+16</span>

d = &#8730;<span style="text-decoration: overline">25</span>

d = 5

For EF

d = &#8730;<span style="text-decoration: overline">(9-5)²+(4-7)²</span>  

d = &#8730;<span style="text-decoration: overline">(4)²+(-3)²</span>

d = &#8730;<span style="text-decoration: overline">16+9</span>

d = &#8730;<span style="text-decoration: overline">25</span>

d = 5

So triangle DEF is also an isosceles triangle as well as a right triangle.

Edwin</pre>