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No trigonometry allowed.

Let $\Delta ABC$ be inscribed inside a circle.Let $P$ be a point on the circle.Let $PD$ and $PE$ be perpendiculars on on $BC$ and $AC$ respectively.Let $DE$ when extended meet $AB$ at $F$. Then it is required to prove that $PF$ is perpendicular to $AB$.

Figure

For convenience of description,I will assume that $P$ is opposite to angle $A$.

I have tried the following things:

  1. Drawing auxiliary lines: Unfortunately, apart from joining $P$ and $C$, $P$ and $A$, $F$ and $P$, and $P$ and $B$, I cannot see any obvious constructions. Moreover,the lines are unnecessarily cluttering up the diagrams.

  2. Using Ptolemy's theorem or it's converse: Note that if $PF$ is indeed perpendicular to $AB$, this would make $BFPD$ a cyclic quadrilateral. Again, it would also make $AFPE$ a cyclic quadrilateral. Unfortunately, we do not have enough information regarding sides and diagonals.

  3. Angle-chasing: Admittedly,this should have been my initial approach, but I was too busy with ratios. Let

    $\angle BAC=x,\angle ACB=y$

    Now we backtrack.If PF is indeed perpendicular to AB,we have that $\angle DPF=180-x-y=\angle ABC$.

    Therefore,our task reduces to proving that $\angle DPF=ABC$

After that,I tried some other constructions(such as extending $FP$) but they turned out to be fruitless attempts.

The facts that I haven't used until now:

  1. The fact that $\Delta ABC$ is inscribed inside a circle.

The only way that I can see a use for this information is by constructing $PB,PA$ and saying that

$\angle BPA=\angle PAC$

But that does not help me get any further.

Some hints as to how I can approach such a problem will be appreciated.

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Can you put a figure? –  Sawarnik Apr 7 at 15:58
1  
@Sawarnik,picture added.Credits to MvG. –  rah4927 Apr 8 at 11:21

3 Answers 3

up vote 1 down vote accepted

Not exactly sure if this is any simpler, or different from whats said from above...but anyways

∠CEP = ∠CDP = 90

Therefore CEDP is a cyclic quad (If an interval subtends equal angles on the same side, then the two end points and the two interval points are concyclic)

Join PB

∠PBF = ∠ACP (The exterior angle at the vertex of a cyclic quad equals to the interior opposite angle) In cyclic quad ACPB

∠ACP = ∠FDP (As above) -In cyclic quad CEDP

*Therefore DPFB is a cyclic quad (If an interval subtends equal angles on the same side, then the two end points and the two interval points are concyclic)

∠PFB + ∠PDB = 180 (Opposite angles in cyclic quad are supplementary)

But PDB = 90 (PD perpendicular to CE)

Therefore:

PF is perpendicular to AF as ∠PFA = 90

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Nice solution.+1. –  rah4927 Apr 10 at 12:00

Clifford’s first circle theorem

By construction, $ABCP$ is cyclic. And since $\angle PDC=\angle PEC=90°$, $CDEP$ is cyclic as well.

Now consider the complete quadrilateral formed by the lines $ABF, ACE, BCD$ and $DEF$. You can view the plane as $\mathbb{CP}^1$, the complex projective line, and in that setup, a line is simply a circle (of infinite radius) passing through the point at infinity. So these four lines are a special case of four circles passing through the point at infinity. Now you can apply Clifford's first circle theorem to conclude that the circles $\bigcirc ABC, \bigcirc AEF, \bigcirc BDF$ and $\bigcirc CDE$ must intersect in a single common point.

Since $\bigcirc ABC$ and $\bigcirc CDE$ intersect in $P$, as seen from the cocircularities stated at the beginning of this answer, you get that $AEFP$ and $BDFP$ must be cyclic as well. Now both of these already have one right angle, so by Thales' theorem, the angle at $F$ must be a right angle as well.

Figure including cocircularities

Angle chasing

Even without projective geometry and Clifford's circle theorems, knowing about these additional cocircularities might help you in making deductions about the various angles. In that sense, this answer here might also serve as a hint. Using that, I was able to find the following angle-chasing sequence. Start by $ABCP$ being cyclic:

\begin{align*} \angle BAP = \angle BCP &= \alpha \\ \angle PAC = \angle PBC &= \beta \\ \angle ACB = \angle APB &= \gamma \\ \angle BAC &= \angle BAP + \angle PAC \\ \angle CBA &= 180° - \angle BAC - \angle ACB \\ \angle CPA &= \angle CBA \end{align*}

$CDEP$ is cyclic as well, and Thales.

\begin{align*} \angle PDC = \angle PEC &= 90° \\ \angle DCP &= \angle BCP \\ \angle DEP &= \angle DCP \\ \angle DEC &= \angle DEP + \angle PEC \\ \end{align*}

Follow these angles some more.

\begin{align*} \angle BDP &= 180° - \angle PDC \\ \angle PBD &= \angle PBC \\ \angle DPB &= 180° - \angle BDP - \angle PBD \\ \angle FBP &= 180° - \angle PBC - \angle CBA \\ \angle FEC &= \angle DEC \\ \angle AEF &= 180° - \angle FEC \\ \angle FAE &= \angle BAC \\ \angle EFA &= 180° - \angle AEF - \angle FAE \\ \angle DFB &= \angle EFA \end{align*}

Then you can check that $\angle DFB = \angle DPB$ so $BDFP$ is circular and therefore Thales.

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Is there any simple proof of Clifford's circle theorem? –  rah4927 Apr 8 at 11:34
    
@rah4927: Not sure. I let my students prove it using the fact that $(A,D;B,E;C,F)$ form a quadrilateral set in the complex plane, which requires a lot of background in projective geometry. With that background, it is easy enough, but obtaining that background takes about one semester so I guess there should be simpler ways. That question for a suitable proof might be worth a post in its own respect, I guess. You could try to obtain a copy of that Penguin Dictionary Wikipedia names as a reference; that might contain a proof or a pointer to one. –  MvG Apr 8 at 11:48
    
,hmm,unfortunately,I have zero experience in projective geometry.Also,which additional co-circularities are you talking about in the last line of your answer? –  rah4927 Apr 8 at 12:00
    
@rah4927: There might be proofs of Clifford's theorems (btw. notice the plural) which don't use projective geometry at all. Not sure if they's use trigonometry instead, but perhaps not. Unless I missed it, $CDEP$ being cocircular was not mentioned in your post. I would hope that if you add lines $AP, BP, CP$ and then consider all the equal angles in all your cyclic quadrilaterals, you might end up with enough conditions to prove your claim. A bit like what I did for this answer I'd hope. –  MvG Apr 8 at 12:10

hint: set $A(x_1,y_1), B(-p,0),C(p,0),O(0,q), P(x_2,y_2), x_1^2+(y_1-q)^2=x^2+(y_2-q)^2=p^2+q^2$

write down $PD\perp BC,PE\perp AC,PF\perp AB$

now you can get $D(x_D,y_D),E(x_E,y_E),F(x_F,y_F)$, prove they are co-line.( there are some heavy work to do.)

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