Question 119374
Quadrilateral ABCD has vertices A(0,-2), B(9,1), C(4,6), and D(1,5). Prove: a) Quadrilateral ABCD is a trapezoid. b) Quadrilateral ABCD is an isosceles trapezoid.

<pre><font size = 4><b>
{{{drawing(400,400,-5,10,-5,10,

graph(400,400,-5,10,-5,10),

line(0,-2,9,1), line(9,1,4,6), line(4,6,1,5),line(1,5,0,-2),
locate(.3,-2,A), locate(9,1,B), locate(4,7,C), locate(1,6,D) )}}}

First we prove that AB and DC are parallel, 
by showing they have the same slope:

Slope of AB
     
m = {{{((1)-(-2))/((9)-(0))}}} = {{{(1+2)/9}}} = {{{3/9}}} = {{{1/3}}}

Slope of DC

m = {{{((6)-(5))/((4)-(1))}}} = {{{1/3}}}

There are both {{{1/3}}} so they are parallel and so ABCD is
a trapezoid.

To show that it is isosceles, we prove that AD = CB

Length of AD

d = {{{sqrt(((1)-(0))^2+((5)-(-2))^2)}}} = {{{sqrt((1)^2+(5+2)^2)}}} = {{{sqrt(1+7^2)}}} = {{{sqrt(1+49)}}}
= {{{sqrt(50)}}} = {{{sqrt(25*2)}}} = {{{5sqrt(2)}}} 

Length of CB

Length of AD

d = {{{sqrt(((9)-(4))^2+((1)-(6))^2)}}} = {{{sqrt((5)^2+(-5)^2)}}} = {{{sqrt(25+25)}}} = {{{sqrt(50)}}} = {{{sqrt(25*2)}}} = {{{5sqrt(2)}}} 

These are both {{{5sqrt(2)}}} so it is also isosceles.

Edwin</pre>