# CGMO 2020: Prove that $X, P, Q, Y$ are concyclic.

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.

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

Here's the hint.

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

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

• Thanks for the hint ! Aug 11 '20 at 4:54

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):

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.

• 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 $X,Y$...) Aug 10 '20 at 19:51

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.

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 $$PQ$$

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

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.

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!

• Looks good to me. Aug 11 '20 at 8:27