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PQRS is a square. The bisector of angle SQR meets PR and SR at T and V respectively. Prove that: QV*TR = QT*SV

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Every angle is trivial to find. Why not use brute force? –  Karolis Juodelė Jan 3 '13 at 12:36
    
@Karolis Juodelė what do you mean by brute force? –  Ghost Jan 3 '13 at 12:37
    
I mean to find the length of every line segment, assuming $PQ = a$. You'll only need a tiny little bit of trigonometry. –  Karolis Juodelė Jan 3 '13 at 12:40
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Oh wait, QTR is similar to SVQ (just write down the angles). The property follows. –  Karolis Juodelė Jan 3 '13 at 12:51
    
oh didn't see that, many thanks. –  Ghost Jan 3 '13 at 12:57
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Hint: As a preliminary step, do some angle-chasing. We can even do it mindlessly, without plan. Just fill in all the angles. The only fact we will need is that the angles of a triangle add up to $180^\circ$.

As a start, note that $\angle RQS=45^\circ$, so $\angle VQS=\angle VQR=22.5^\circ$. But $\angle QSV=45^\circ$, so $\angle SVQ=180^\circ-(22.5^\circ+45^\circ)=112.5^\circ$. You may find things less messy if you let $22.5^\circ=a$. Then the angles of $\triangle QSV$ are $a$, $2a$, and $5a$.

Keep filling in angles. When you are finished, or earlier, look for similar triangles. You will notice that $\triangle QSV$ has the same angles as $\triangle QRT$. So these two triangles are similar.

Your desired equality is a direct consequence of this similarity. It comes more naturally as $\frac{SV}{QV}=\frac{TR}{QR}$. Or perhaps note that $SV$ and $TR$ are corresponding sides of our two similar triangles, as are $QV$ and $QT$, so $\frac{SV}{TR}=\frac{QV}{QT}$.

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