Thread: Prove that PQR is isosceles?

1. I can't figure this one out...(I may not have drawn it exactly how it was drawn in the book)

2. Originally Posted by woof123
I can't figure this one out...(I may not have drawn it exactly how it was drawn in the book)
angle PRQ = 180 - 2x by supplementary angles.

Thus angle PQR = 180 - (x) - (180 - 2x) = x. What does this tell you?

-Dan

OK, I get x, meaning the sides opposite to the x angles are equal.

Thanks so much

4. Okay, let's label the unknown angles:

Now, we know the following:

$\displaystyle 2x+\gamma=180^{\circ}$

$\displaystyle x+\alpha+\gamma=180^{\circ}$

$\displaystyle 2x+\beta+\delta=180^{\circ}$

$\displaystyle x+\alpha+\beta+\delta=180^{\circ}$

We need to show that:

$\displaystyle x=\alpha$

Look at the first two equations above, and we see that:

$\displaystyle 180^{\circ}-\gamma=2x=x+\alpha$

What does this imply?

5. I would like to express my opinion here

Theorem: The exterior angle formed when a side of a triangle is produced
is equal to the sum of the two interior opposite angles.

Now using this theorem,

$2x=x+\angle PQR$
$2x-x= \angle PQR$
$x= \angle PQR$

Then we may recall the theorem,

Theorem (Converse of the theorem on isosceles triangles):
The sides opposite equal angles of a triangle are equal

Now since we have proved that $\angle PQR =\angle QPR$ what can be said about the sides PQ and PR ?