Prove that $ST$ is antiparallel with $PQ$ Given is $\triangle PQR$ with altitudes $PS$ and $QT$. Prove that $ST$ is antiparallel with $PQ$.
What I've done up until now:


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*Given: $\triangle PQR$ with altitudes $QT$ and $PS$

*What to prove: $P_1 = S_2$ (click here for the image)
I want to prove this with extremely basic mathematics, because my book solves this using the reverse of Thales' theorem, and cyclic quadrilaterals and such, and I believe there must be a simpler way to prove this, I however can't figure out how. Is there a simpler way?
 A: 
Looking at the figure above, we need to prove that $$\delta = \beta + \gamma.$$
From the figure we can deduce that $\triangle RSP \sim \triangle RTQ$ and
$\triangle SOQ \sim \triangle TOP$.
So we can conclude that
$$\frac{a}{a'}=\frac{b}{b'}$$
Let's rearrange that proportion in this way:
$$\frac{a}{b}=\frac{a'}{b'}$$
But by the picture
$\angle QOP = \angle SOT$ (vertically opposite angles).
Therefore
$$\triangle POQ \sim \triangle TOS$$
Now we can conclude that
$$\beta = \mu$$ and
$$\alpha = \theta.$$
Applying the exterior angle theorem in $\triangle TQS$ we get:
$$\delta = \mu + \gamma \Rightarrow$$
$$\Rightarrow \delta = \beta+\gamma$$
Therefore $ST$ and $PQ$ are anti-parallel with respect to $RP$ and $RQ$.
A: Using the picture from Ricardo's answer.
Draw a circle of diameter PQ. Since $PTQ =90^\circ$, $T$ is a point on this circle. $PSQ =90^\circ$, $S$ is a point on this circle.
Thus $PQST$ are on the same circle, and hence $\angle TPQ + \angle TSQ =180^\circ =\angle RST + \angle TSQ$$
