I can't figure this one out...(I may not have drawn it exactly how it was drawn in the book)
I can't figure this one out...(I may not have drawn it exactly how it was drawn in the book)
OK, I get x, meaning the sides opposite to the x angles are equal.
Thanks so much
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?
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 ?