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Given is $\triangle PQR$ with altitudes $PS$ and $QT$. Prove that $ST$ is antiparallel with $PQ$.

What I've done up until now:

  • 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?

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    Sorry, my drawing is incorrect. $ \angle P_{12} $ must be equal to $\angle S_2$2012-12-08

2 Answers 2

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Triangle PQR and two heights 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 $RQ$.

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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$$