6
$\begingroup$

In the quadrilateral $ABCD$, $AB=AD$, $CB=CD$, $\angle ABC =90^\circ$. $E$, $F$ are on $AB$, >$AD$ and $P$, $Q$ are on $EF$($P$ is between $E, Q$), satisfy $\frac{AE}{EP}=\frac{AF}{FQ}$. $X, Y$ are on $CP, CQ$ that satisfy $BX \perp CP, DY \perp CQ$. Prove that $X, P, Q, Y$ are concyclic.

My Progress: Couldn't proceed much . I noted that $ABCD$ is cyclic quad with diameter $AC$ . I feel to use POP on C , so it is enough to show that $CX\cdot CP= CY\cdot CQ$ . But I am not sure about how to use "$\frac{AE}{EP}=\frac{AF}{FQ}$" criteria .

Please post hints rather than solution. It really helps me a lot.

Thanks in advance.

$\endgroup$
4
  • $\begingroup$ 1) A picture would be useful. 2) I'd try vectors approach $\endgroup$ Aug 10 '20 at 15:28
  • $\begingroup$ If $AB=BC=CD$ and $ABC$ is $90$ degree, then $ABCD$ is a square. I guess there is a typo? $\endgroup$
    – cr001
    Aug 10 '20 at 15:29
  • $\begingroup$ @AlexeyBurdin , actually picture for this is really hard to draw because of the ratios given $\endgroup$ Aug 10 '20 at 15:33
  • $\begingroup$ @cr001 , yes it was a typo , I meant AB=AD $\endgroup$ Aug 10 '20 at 15:33
4
$\begingroup$

Here's the hint.

($1$) The colored line are of importance. Think what the color means.

($2$) Make use of parallel line ratio.

enter image description here

$\endgroup$
1
  • $\begingroup$ Thanks for the hint ! $\endgroup$ Aug 11 '20 at 4:54
4
$\begingroup$

This is a full proof following the natural wish in the OP to use the power of the point $C$ w.r.t. the points that should be on the circle.


The picture first (and try to figure out a property of the line $P'Q'$ without further reading):

Mathematics stackexchange question 3786317

Here, many elements are needed only for having a faithful picture. The points needed in the proof are the red ones:

  • $\color{red}Z$ is the intersection of the lines $EPQF$ and $AC$,

  • $\color{red}{P'}$ is $AB\cap CX$, and $\color{red}{Q'}=AD\cap CY$.

We compute $CX\cdot CP$, trying to express it in a "symmetricaly way" w.r.t. the given symmetry of the figure. First, since there is a right angle in $B$ in $\Delta BCP'$ we have $$ CB^2= CX\cdot CP'\ . $$ So it is natural to try to deal with the proportion $CP:CP'$ or with some derivated form of it.


A further hint so far:

Using for instance for the equality marked $(!)$ below the sine theorem in $\Delta AEZ$ and $\Delta AFZ$ we get: $$ \tag{$1$} \frac {PE}{QF}= \frac {AE}{AF}\overset{(!)}{=\!=} \frac {ZE}{ZF}= \frac {ZP}{ZQ}\ . $$


Lemma: $$ \tag{$2$} \color{red}{P'Q'}\|EF\ . $$ Proof: Menelaos in $\Delta EAZ$ for the "secant" line $CPP'$, respectively in $\Delta FAZ$ for the "secant" line $CQQ'$ gives: $$ \begin{aligned} 1&= \frac{PZ}{PE}\cdot \color{blue}{\frac{P'E}{P'A}}\cdot \frac{CA}{CZ} \ , \\ 1&= \frac{QZ}{QE}\cdot \color{blue}{\frac{Q'E}{Q'A}}\cdot \frac{CA}{CZ} \ , \end{aligned} $$ and the middle blue proportions are equal, since the others are correspondingly. (Use $(1)$.) Thus the claimed parallelism.

$\square$


The finish is now: $$ \begin{aligned} \frac {CX\cdot CP}{CY\cdot CQ} &= \frac {CX\cdot CP'}{CY\cdot CQ'}\qquad\text{ since }PQ\|P'Q' \\ &= \frac {CB^2}{CD^2} =1\ . \end{aligned} $$ $\square$


Note: The green region suggests that we are trying to "move proportions" from the line $CPP'$ to the line $CZA$ by using conveniently triangles "based" on the one or the other line.

$\endgroup$
1
  • $\begingroup$ In the picture, the line $BX$ is with bad glasses going through $Y$, this is not the case in the hi res geogebra picture i exported... (I was needing a good position for $P,Q$, but this was breaking the position for $X,Y$...) $\endgroup$
    – dan_fulea
    Aug 10 '20 at 19:51
4
$\begingroup$

Let $EF$ cut $AC$ at $R$. Then

  • $AR$ is angle bisector for $\angle EAF$ so ${AE\over AF} = {ER\over RF}$ and thus $${EP\over PR} ={FQ\over QR}\;\;\;(*)$$
  • Reflect $E,P$ and $X$ across $AC$, we get $E',P'$ and $X'$. Because of $(*)$ we have $E'F||P'Q$ and $Y,X',C,D$ are concyclic.
  • Let $\angle CDX'= \phi$, then $\angle CYX' = \phi $ and $\angle X'DA = 90-\phi$, so $\angle QYX' = 180-\phi $ and $\angle X'P'Q = \phi$ which means $X',Y , Q$ and $P'$ are concyclic.
  • By PoP with respect to point $C$ we see that $P,X,Y,Q$ are conyclic.

enter image description here

$\endgroup$
3
$\begingroup$

enter image description here

This is a figure of the given situation in Geogebra.

Hint: We get $P'$ and $Q'$ by rotating $P$ and $Q$ about $E$ and $F$ respectively. Hence, we have $EP=EP'$ and $FQ=FQ'$.

Since, it is given that $\displaystyle\frac{AE}{PE}=\frac{AF}{FQ}$, line $P'Q'$ is parallel to $PQ$

$\endgroup$
2
  • $\begingroup$ I didn't understand about the rotation part..can you explain it once ? $\endgroup$ Aug 11 '20 at 5:07
  • $\begingroup$ $P$ is rotated about $E$ such that the rotated point $P'$, $A$ and $E$ are collinear. Same for $Q$. Then use inverse BPT. $\endgroup$
    – SarGe
    Aug 11 '20 at 7:25
2
$\begingroup$

Thanks everyone, for their hints! I think I got the solution (using @cr001 's hint ) . I hope someone can verify this proof .

I am going to use @cr001's diagram.

enter image description here

Let $AC\cap EF= I$ . Let $H_1$ be the foot of the perpendicular from $P$ to $BC$. Let $H_2$ be the foot of the perpendicular from $Q$ to $DC$. Let $H_3$ be the foot of the perpendicular from $P$ to $BA$.Let $H_4$ be the foot of the perpendicular from $Q$ to $AD$.Let $H_5$ be the foot of the perpendicular from $I$ to $BA$.Let $H_6$ be the foot of the perpendicular from $I$ to $AD$.

Now note that AI is the angle bisector of EF. So we have $\frac {AE}{AF}=\frac {EI}{IF}$ (using the angle bisector theorem )

Also we have $\frac {EP}{EI}=\frac {PH_3}{IH_5}$ (using similarity ).

similarly, we have $\frac {FQ}{FI}=\frac {QH_4}{IH_6}$ (using similarity ).

So we have $\frac {IH_5}{PH_3} \cdot\frac {QH_4}{IH_6}= \frac {EI}{EP}\cdot\frac {FQ}{FI}=\frac {AE}{EP}\cdot \frac {FQ}{AF}=1 \implies \frac {IH_5}{PH_3} \cdot\frac {QH_4}{IH_6}=1 \implies QH_4= PH_3$ (since $IH_5=IH_6$).

So we have $DH_2=QH_4= PH_3=BH_1 \implies CH_1=CH_2$ .

Now, since $\angle PH_1B=\angle BXP=90 $, we get $PH_1BX$ cyclic .

Similarly $QYH_2D$ is cyclic.

So $\Bbb P(C,(PH_1BX))= CH_1\cdot CB=CH_2\cdot CD=\Bbb P(C,(PH_1BX))$

So $\Bbb P(C,(PH_1BX))=\Bbb P(C,(QH_2YD)) \implies CX \cdot CP=CY \cdot CQ \implies XYPQ$ is cyclic .

And we are done!

$\endgroup$
1
  • 2
    $\begingroup$ Looks good to me. $\endgroup$
    – cr001
    Aug 11 '20 at 8:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.