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ABCD is a square. F is a point on BD. CF bisects angle ACD. Q lies on CD. BQ is perpendicular to CF. If AC and BD intersect at E and BQ intersects AC at P, then prove that DQ = 2(PE).

I tried to use Midpoint theorem, but anyhow obstacles come on my way. Can anyone help?

Please do provide an euclidean approach, thank you.

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  • $\begingroup$ There are multiple theorems that have been called the Midpoint Theorem. Which one do you mean? $\endgroup$ – Rory Daulton Oct 18 '15 at 11:59
  • $\begingroup$ The one you linked just now, sir. $\endgroup$ – Swapnil Das Oct 18 '15 at 12:02
  • $\begingroup$ But there are several linked to in that search page. Do you mean the featured one at the top? $\endgroup$ – Rory Daulton Oct 18 '15 at 12:03
  • $\begingroup$ Ya, you are right :) $\endgroup$ – Swapnil Das Oct 18 '15 at 12:04
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I'm writing this answer referring to Rory's diagram.

First off apply Menelaus theorem in $\triangle CED$ taking $BQ$ as the transversal. This will give you $$\frac{PE}{DQ}=\frac 12 \cdot \frac {PC}{QC}$$

Now since $\triangle CRP \cong \triangle CRQ$ we get $PC=QC$ and the result immediately follows.

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Here is a coordinate approach. In this diagram I have taken the side of the square to have length one, and points $A,B,C,D$ have the obvious coordinates.

enter image description here

Using simple analytic geometry we get these facts:

  • $\overleftrightarrow{AC}$ has the equation $y=x$.
  • $\overleftrightarrow{BD}$ has the equation $y=1-x$.
  • $E$ is the point $(\frac 12,\frac 12)$.
  • $\measuredangle ACD=45°$.
  • $\measuredangle DCF=22.5°$.
  • $\overleftrightarrow{CF}$ has the equation $y=(\sqrt 2-1)x$.
  • $\overleftrightarrow{BQ}$ has the equation $y=1-(\sqrt 2+1)x$.
  • $Q$ is the point $(\sqrt 2-1,0)$.
  • $P$ is the point $(1-\frac{\sqrt 2}2,1-\frac{\sqrt 2}2)$
  • $DQ=2-\sqrt 2$.
  • $PE=1-\frac{\sqrt 2}2$.
  • $DQ=2\cdot PE$.

Let me know if you need the details on any of those points. Most are quite simple: the most difficult is finding $PE$.

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    $\begingroup$ Cool, I liked it. I hope you could provide an Euclidean approach too. $\endgroup$ – Swapnil Das Oct 18 '15 at 12:41
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    $\begingroup$ @SwapnilDas: No, this was not a hard question. The hardest part was preparing the diagram and deciding which facts to include and in what order. I don't have the time for a Euclidean approach: I'm off to church then to a movie date with my wife. $\endgroup$ – Rory Daulton Oct 18 '15 at 12:46
  • $\begingroup$ Aahh.. Sure. Enjoy! $\endgroup$ – Swapnil Das Oct 18 '15 at 12:47
  • $\begingroup$ What is this drawn with? $\endgroup$ – Eric Oct 18 '15 at 16:14
  • $\begingroup$ @Eric: Geogebra, which I highly recommend and is often advertised on this site. (I'm back now, and I see someone give a Euclidean proof, so I can concentrate on school work.) $\endgroup$ – Rory Daulton Oct 18 '15 at 19:36

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