MHB Is Triangle PQR Isosceles Based on Angle Equality?

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Triangle PQR is determined to be isosceles based on the equality of angles derived from supplementary angle relationships. The calculations show that angle PQR equals angle QPR, both represented as x. This equality implies that the sides opposite these angles, PQ and PR, are equal. The discussion references the theorem that states sides opposite equal angles in a triangle are equal, confirming the isosceles nature of triangle PQR. Thus, the conclusion is that triangle PQR is indeed isosceles.
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I can't figure this one out...(I may not have drawn it exactly how it was drawn in the book)

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woof123 said:
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
 
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:

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

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

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

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

We need to show that:

$$x=\alpha$$

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

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

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 ?
 
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