Is Triangle PQR Isosceles Based on Angle Equality?

[woof123]

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

View attachment 6270

 

[topsquark]

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

 

[woof123]

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

Thanks so much

 

[MarkFL]

Okay, let's label the unknown angles:

\begin{tikzpicture}
\draw[blue,thick] (0,0) -- (12,0);
\draw[blue,thick] (12,0) -- (8,4);
\draw[blue,thick] (8,4) -- (0,0);
\draw[blue,thick] (8,4) -- (6,0);
\node[left=5pt of {(0,0)}] {\large P};
\node[above=5pt of {(8,4)}] {\large Q};
\node[below=5pt of {(6,0)}] {\large R};
\node[right=5pt of {(12,0)}] {\large S};
\node[right=15pt of {(0,0)},yshift=5pt] {\large $x$};
\node[right=5pt of {(6,0)},yshift=7pt] {\large $2x$};
\node[left=5pt of {(8,4)},yshift=-13pt] {\large $\alpha$};
\node[below=3pt of {(8,4)}] {\large $\beta$};
\node[left=10pt of {(12,0)},yshift=7pt] {\large $\delta$};
\node[above=0pt of {(6,0)},xshift=-3pt] {\large $\gamma$};
\end{tikzpicture}

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?

 

[mathlearn]

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 ?