Using homothety,P is the center,A ->C B ->D Q ->Q’ Then we have Q’,P,Q are collinear by homothety ∠AQ’D= ∠CQB=∠AQD => AQ’QD is concyclic => ∠DAQ=∠DQ’Q=∠DQ’P=∠BQP

Nice short solution , plane geometry! Will you please resend your soln. with a diagram. Moreover, I didn't understand what you wanted to mean by A ->C B ->D Q ->Q’ . proabably you meant constructions!!!

Sunkern_sunflora wrote: Nice short solution , plane geometry! Will you please resend your soln. with a diagram. Moreover, I didn't understand what you wanted to mean by A ->C B ->D Q ->Q’ . proabably you meant constructions!!! Sorry，I didn't say it clearly A ->C means A,C are homothety w.r.t P

plane geometry wrote: Quote: Sorry，I didn't say it clearly A ->C means A,C are homothety w.r.t P What on the earth does homothety mean????

Sunkern_sunflora wrote: plane geometry wrote: Quote: Sorry，I didn't say it clearly A ->C means A,C are homothety w.r.t P What on the earth does homothety mean???? SEE HERE http://mathworld.wolfram.com/Homothecy.html

Thank you for providing me with that link plane geometry! I will be very obliged if you kindly clarify , cause I couldn't make out much about homothecy from your link.

I have a nice and easy solution with trigonometry. Have fun!

plane geometry wrote: Using homothety,P is the center,A ->C B ->D Q ->Q’ Then we have Q’,P,Q are collinear by homothety ∠AQ’D= ∠CQB=∠AQD => AQ’QD is concyclic => ∠DAQ=∠DQ’Q=∠DQ’P=∠BQP that is actually the official solution, isn't it? I found a solution without even mentioning the word homothety, but it is a bit unnatural.

April wrote: The diagonals of a trapezoid $ ABCD$ intersect at point $ P$. Point $ Q$ lies between the parallel lines $ BC$ and $ AD$ such that $ \angle AQD = \angle CQB$, and line $ CD$ separates points $ P$ and $ Q$. Prove that $ \angle BQP = \angle DAQ$. Author: unknown author, Ukraine ___________________ My answer without words

Just out of curiosity, what was the motivation behind this solution? How would one know to construct such a point Q'?

Hi, does anyone have any non-homothetic solutions to this problem? will be greatly appreciaited. Best wishes, manifest desniyeeeeeeee

I found pretty weird solution using fourth lemma from here: http://sem.edu.pl/materialy/sem-plakat3.pdf We apply this lemma to triangle $BAQ$ (constructing earlier new point) and solution from now is obvious. But unfortunately this lemma isn't so obvious. I'm not aware of its proof without using trigonometric Ceva's theorem and law of sines :/.

$P$ is the inner center of similitude of the circles $QAD$ and $QBC$. After noticing this fact, the rest part of the problem becomes very easy. This proof is almost same as the previous proof by homothety. But I posted this one because viewing from this perspective provides a better excuse to extend $QP$. (The motivation of what SnowEverywhere was talking.) SnowEverywhere wrote: Just out of curiosity, what was the motivation behind this solution? How would one know to construct such a point Q' ?

problem is easy by ceva theorem.

A polar transformation with centre Q and an arbitary radius solves it immediately

Another consequence of the infamous, forgotten isogonality Lemma, just draw the parallel from $Q$ and boom, bulls eye!

This isn't too hard to complex bash either; by setting $q=0$ and fixing $b,c$, one gets that for some $z$, $b=\frac{cz^2(c-\overline{c})}{cz^2-\overline{c}}$ and $a=\frac{dz^2(d-\overline{d})}{dz^2-\overline{d}}$. Moreover, $p = \frac{cd-ab}{c+d-a-b}$. Now one needs to check that $\frac{d-a}{-a}:\frac{b}{p}$ is real, which is not hard.

After casually leaving this unfinished for a year:

Let $K$ be the point on $PQ$ such that $BQ \parallel DK$. It's easy to see that $ADK$ and $CBQ$ are homothetic at $P$, implying $$\angle AKD = \angle CQB = \angle AQD$$which means $ADQK$ is cyclic. Hence, $$\angle DAQ = \angle DKQ = \angle BQP$$as desired. $\blacksquare$

over the past couple of days. Thus, it's quite strange that I didn't manage to spot it here.

[asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair A = (-34.42819,67.89340); pair B = (-74.88348,-31.82015); pair C = (25.84231,-32.05941); pair D = (8.33576,67.79182); pair Q = (74.46684,-13.21240); pair P = (-16.46590,38.10467); pair X = (100.00792,-32.23557); import graph; size(15cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); pen wwzzqq = rgb(0.4,0.6,0); draw(arc(Q,17.37791,129.22787,143.32099)--Q--cycle, linewidth(0.75) + blue); draw(arc(Q,17.37791,-172.89803,-158.81352)--Q--cycle, linewidth(0.75) + blue); draw(arc(A,21.23966,-36.67900,-0.13609)--A--cycle, linewidth(0.75) + red); draw(arc(Q,21.23966,150.56218,187.10196)--Q--cycle, linewidth(0.75) + red); draw(arc(X,21.23966,143.32099,179.86390)--X--cycle, linewidth(0.75) + red); draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--D, linewidth(0.5)); draw(D--A, linewidth(0.5)); draw(B--D, linewidth(0.5)); draw(A--C, linewidth(0.5)); draw(D--Q, linewidth(0.5) + red); draw(Q--C, linewidth(0.5) + red); draw(A--Q, linewidth(0.5) + wwzzqq); draw(Q--B, linewidth(0.5) + wwzzqq); draw(Q--P, linewidth(0.5) + blue); draw(arc(A,21.23966,-36.67900,-0.13609), linewidth(0.75) + red); draw(arc(A,19.11570,-36.67900,-0.13609), linewidth(0.75) + red); draw(arc(Q,21.23966,150.56218,187.10196), linewidth(0.75) + red); draw(arc(Q,19.11570,150.56218,187.10196), linewidth(0.75) + red); draw(Q--X, linewidth(0.5)); draw(X--C, linewidth(0.5)); draw(arc(X,21.23966,143.32099,179.86390), linewidth(0.75) + red); draw(arc(X,19.11570,143.32099,179.86390), linewidth(0.75) + red); draw((-106.73860,-12.78198)--(143.20593,-13.37567), linewidth(0.5) + blue); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, dir(270)); dot("$D$", D, NE); dot("$Q$", Q, NE); dot("$P$", P, dir(180)); dot("$X$", X, SE); [/asy][/asy] Let $X=PQ\cap BC$. Also $\ell$ denote the line through $Q$ parallel to $BC$. Now by DDIT on quadrangle $ACBD$ w.r.t. $Q$, we get that $\left\{(QP,Q\infty_{BC}),(QA,QB),(QC,QD)\right\}$ are pairs under some Involution. Now due to our given angle condition, we get that $\angle AQB$ and $\angle DQC$ share the same angle-bisectors. This gives us that $\left\{(QD,QC),(QA,QB)\right\}$ are reciprocal pairs under the Involution of isogonal conjugation w.r.t. the bisector of $\angle QAB$. This thus also means that $QP$ and $\ell$ are isogonal w.r.t. the angle-bisector of $\angle QAB$. Thus to finish, we have that, \begin{align*} \measuredangle DAQ&=\measuredangle DAX\\ &=\measuredangle CXA\\ &=\measuredangle (\ell,QA)\\ &=\measuredangle (\ell,QP)+\measuredangle PQA\\ &=\measuredangle (\ell,QP)+\measuredangle (BQ,\ell)\\ &=\measuredangle BQP ,\end{align*}and we are done.

This is a boring problem :unamused: it's one construction which trivializes, nothing to learn from it. Let the line through D parallel to BQ intersect PQ at R. It's obvious by sim triangle ratio of sides that ADR and CBQ are homothetic at P (RP/RQ=RD/RB=RA/RC). Then ARD=CQB=AQD, meaning R lies on the circumcircle of ADQ. Then DAQ=DRQ=BQP, which finishes. $\blacksquare$

Cute! Suppose the homothety at $P$ sending $BC$ to $DA$ takes $Q$ to point $R.$ Then we have $BQ \parallel DR,$ and $\angle ARD=\angle CQB=\angle AQD,$ so $ARQD$ is cyclic. Then, we have $\angle BQP=\angle BQR=\angle DRQ=\angle DAQ$ and we are done.

Let $X$ be the second intersection of line $PQ$ with $(BQC)$. Clearly $P$ is the center of the homothety from $(BQC)$ to $(AQD)$. This homothety sends arc $BX$ in $(BQC)$ to arc $QD$ in $(AQD)$, hence $\angle BQX = \angle DAQ$.

Extend $QP$ to meet the circumcircle of $ADQ$ at a point $R$. Then it is easy to see from the angle condition that there exists a homothety taking $\triangle BCQ \mapsto \triangle DAR$ so we have $RD \parallel BQ$ from which we finish as $\angle QAD = \angle QRD = \angle BQP$.

Take a homothety centered at $P$ that maps $C$ to $A$ and $B$ to $D$, because they are similar triangles. Let us say this homothety sends $\triangle{BCQ}$ to $\triangle{DAT}$. Then: \[\angle DTA=\angle BQC=\angle AQD,\]so $ADQT$ is cyclic. We then angle chase to see: \[\angle BQP=\angle DTQ=\angle DAQ,\]as desired $\square$

ok heres th e dumb solution first i guess By $DDIT$ involution swaps $(QA,QB),(QC,QD),(QP,Q\infty_{AD})$ so we're done by angle chase.

Homothety at $P$ sending $AD$ to $CB$ will also send $Q$ to a point $R$ such that $AQ \parallel CR$ and $BQ \parallel DR$. Thus \[\angle DRA = \angle BQC = \angle DQA \implies ADQR \text{ cyclic} \implies \angle BQP = \angle QRD = \angle QAD. \quad \blacksquare\]

Assume WLOG that $\frac{AD}{BC} = r \le 1,$ and let $Q'$ be the image of $Q$ under a homothety $h$ centered at $P$ with scale $-r.$ Then $h(A) = C$ and $h(D) = B,$ so the given condition is equivalent to $ADQQ'$ being cyclic. This means that $\angle DAQ = \angle DQ'Q.$ Therefore, we'd like to show that $\angle BQQ' = \angle DQ'Q,$ or $DQ' \parallel BQ.$ This is evident since $Q \to Q'$ and $D \to B.$

Let $\mathcal{H}$ be the homothety sending $AD \to BC$, centered at $P$. Then let $\mathcal{H}(Q) = X$. Notice that $\triangle{BCQ} \sim \triangle DAX$, by homothety. It follows that $\angle{AQD} = \angle BQC = \angle{DXA}$, so $ADQX$ is cyclic. From our homothety, we also know that $BQ \parallel XD \implies \angle DAQ = \angle QXD = \angle BQP$, so we are done. Nice problem. I think the solution makes it look easy but it's a bit hard to spot the homothety centered at $P$.

If $Q'$ is the image of $Q$ under the homothety (centered at $P$) sending $\overline{AD}$ to $\overline{CB}$ then the angle condition implies $BCQQ'$ cyclic and $\measuredangle BQP=\measuredangle BQQ'=\measuredangle BCQ'=\measuredangle DAQ$. $\blacksquare$ Remark: I think this problem becomes 10x (not an exaggeration) easier if you make $ABCD$ a self-intersecting trapezoid, i.e. such that $ADBC$ is convex instead of $ABCD$. If we translate the given angle conditions to the appropriate directed angle equalities (by drawing the diagram for when $ABCD$ is convex and inspecting orientations), the problem should still be true when $ADBC$ is convex. It's generally much easier to spot positive homotheties than negative ones, and in this case it becomes much easier (maybe even obvious) to notice that $\overline{AD} \mapsto \overline{CB}$ gives you a cyclic quadrilateral. In fact, when $P$ goes to infinity in this case I think the resulting problem is quite well-known (and older than this).

Using directed angles, the condition is $\measuredangle AQD = \measuredangle CQB$, and we need to show that $\measuredangle BQP = \measuredangle DAQ$. Take a pole-polar duality at $Q$ with arbitrary radius. Denote by $l_P$ the polar of $P$, $P_l$ the pole of $l$. Since $BC//AD$, $P_{BC}, Q, P_{AD}$ are collinear. Also, $l_P=P_{AC}P_{BD}$. $\measuredangle AQD = \measuredangle CQB \implies \measuredangle(l_A,l_D)=\measuredangle(l_C,l_B) \implies \measuredangle P_{AC}P_{AD}P_{BD} = \measuredangle P_{AC}P_{BC}P_{BD} \implies P_{AC}, P_{AD}, P_{BC}, P_{BD}$ are concyclic. Thus, $\measuredangle P_{BC}P_{BD}P_{AC} = \measuredangle P_{BC}P_{AD}P_{AC} = \measuredangle QP_{AD}P_{AC} \implies \measuredangle(l_B,l_P) = \measuredangle(\overline{AD},l_A) \implies \measuredangle BQP = \measuredangle DAQ$.

wow homothety is so nice but imagine being able to find that The condition is essentially saying that $QB$ and $QA$ are isogonal in $\triangle QCD$. The key idea is to that since we need to understand how $QP$ divides up the angle $\angle BQA$, we use Trig Ceva on $\triangle QAB$. We have by Trig Ceva that $$\frac{\sin\angle BQP}{\sin\angle AQP}=\frac{\sin\angle QBD}{\sin\angle ABD}\cdot \frac{\sin\angle CAB}{\sin\angle QAC}$$. However, we have by law of sines on $\triangle BQD$ and $\triangle BAD$ that $$\frac{\sin\angle QBD}{\sin\angle ABD}=\frac{QD}{AD}\cdot\frac{\sin\angle BQD}{\sin\angle BAD}$$and similarly $$\frac{\sin\angle CAB}{\sin\angle QAC}=\frac{BC}{QC}\cdot\frac{\sin\angle ABC}{\sin\angle CQA}.$$However note that $\sin\angle ABC=\sin\angle BAD$ because $BC\parallel AD$, and $\angle BQD=\angle CQA$ by the isogonality of $QB$ and $QA$. Thus, we simply have $$\frac{\sin\angle BQP}{\sin\angle AQP}=\frac{QD}{AD}\cdot \frac{BC}{QC}=\frac{\sin\angle QAD}{\sin\angle AQD}\cdot\frac{\sin\angle CQB}{\sin\angle CBQ}=\frac{\sin\angle QAD}{\sin\angle CBQ}.$$ Note that $$\angle BQA=180-\angle QBA-\angle QAB=\angle CBQ+\angle DAQ,$$so we also have $\angle BQP+\angle AQP=\angle QAD+\angle CBQ$, and because of $$\frac{\sin\angle BQP}{\sin\angle AQP}=\frac{\sin\angle QAD}{\sin\angle CBQ},$$we have $\angle BQP=\angle QAD$ and $\angle AQP=\angle CBQ$, as desired.

Apply DDIT from $Q$ to $ADBC$ to get involution pairs $(\overline{QP},\overline{Q\infty})$; $(\overline{QA},\overline{QB})$; $(\overline{QC},\overline{QD})$. The last two pairs tell us the involution is an isogonality and just for convenience define $T=\overline{Q\infty} \cap \overline{AB}$, just look at this simple angle chase \[\angle BQP=\angle TQA=\angle DAQ\]and we are done.