Since $P$ defines $Q$ unambiguously, then it's enough to prove the converse, i.e. that the relation holds if $\odot(APQ)$ is tangent to $\odot(BOC).$ Let $D,E,F$ be the midpoints of $BC,CA,AB.$ $(N) \equiv \odot(DEF)$ and $(K) \equiv \odot(OBC).$ It's known that $AK,AN$ are isogonals WRT $\angle BAC,$ thus the inversion $(A,\sqrt{AB \cdot AE})$ followed by symmetry WRT the angle bisector of $\angle BAC$ swaps $(N)$ and $\odot(OBC).$ Thus $\odot(APQ)$ goes to the tangent $P'Q'$ of $(N)$ antiparallel to $BC$ that leaves $A,N$ on a same side $\Longrightarrow$ $P'Q'$ is the tangent of $(N)$ at $D.$ So denoting $B_{\infty}$ and $C_{\infty}$ the points at infinity of $AC,AB,$ by involution properties, we get then $(P,B,A,C_{\infty})=(P',E,B_{\infty},A)$ and $(Q,C,A,B_{\infty})=(Q',F,C_{\infty},A).$ But $(P',E,B_{\infty},A)=(Q',C_{\infty},F,A)=(Q',F,C_{\infty},A)^{-1}$ $\Longrightarrow$ $(P,B,A,C_{\infty}) \cdot (Q,C,A,B_{\infty}) =1$ $\Longrightarrow$ $BP \cdot CQ=AP \cdot AQ.$ Remark: From this proof we also get that the tangency point between $\odot(APQ)$ and $\odot(OBC)$ lies on the A-symmedian of $\triangle ABC.$

Dear Mathlinkers, this is just an observation... Sometimes when a relation is given between two points in a triangle, we forget to present a geometrical construction of these two points... If yes, we will observed that we can have more than a solution... Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, this is just an observation... Sometimes when a relation is given between two points in a triangle, we forget to present a geometrical construction of these two points... If yes, we will observed that we can have more than a solution... Sincerely Jean-Louis It's not hard to find the construction of $ P $ and $ Q $ . From the condition (the center of $ \odot (APQ) $ lie on A-altitude) we know $ PQ $ is anti-parallel to $ BC $ WRT $ \angle A $ . From $ BP \cdot CQ=AP \cdot AQ $ we get there exist a point $ R $ on $ BC $ such that $ RP \parallel AC $ and $ RQ \parallel AB $ . Notice that $ AR $ pass through the midpoint of $ PQ $ , so $ AR $ is A-symmedian of $ \triangle ABC $ , hence we get the construction of $ P $ and $ Q $ as following : 1. Let $ R $ be the intersection of A-symmedian of $ \triangle ABC $ with $ BC $ . 2. Let $ P \in AB, Q \in AC $ such that $ RP \parallel AC, RQ \parallel AB $ . ______________________________________________________________________________________ Actually, the tangent point of $ \odot (OBC) $ and $ \odot (APQ) $ is the projection of $ O $ on A-symmedian of $ \triangle ABC $

Dear, Thank for your precise answer correponding to the problem... I was just speaking about only on the relation BP.CQ = AP. AQ which conduct more than one solution par P and Q... Now with the restriction of the problem, it is OK. Sincerely Jean-Louis

Let $X$ be point such that $APXQ$ is an parallelogram $\Rightarrow \frac{PB}{PX}=\frac{QX}{QC} \Rightarrow \triangle BPX \sim \triangle XQC \Rightarrow \angle BXP= \angle XCQ \Rightarrow B,X$ and $C$ are collinear. Easy see that $\angle APQ = \angle BCA \Rightarrow BPQC$ is cyclical, If $Z$ is Miquel point of $\triangle ABC$ (Points $P,Q,X$) then $2\angle BAC=\angle BZC=\angle BOC$ and $\angle BXP=\angle BZP=\angle BCA=\angle QPA=\angle PQX =\angle PQZ + \angle ZCB$ Then $Z$ is tangency point of $\odot (BOC)$ and $\odot (PAQ)$

Similar problem Source: ONEM - Peru 2014, problem 4 - category 3 Let $ABC$ be an acute-angle triangle with circumcenter $O$, if $D,E,F$ lies on the sides $BC, CA$ and $AB$ respectively, such that $BDEF$ is a parallelogram. Suppose that $DF^2=AE.EC < \frac{AC^2}{4}$. Prove that $\odot (AOC)$ and $\odot (DBF)$ are tangents

drmzjoseph wrote: Let $X$ be point such that $APXQ$ is an parallelogram $\Rightarrow \frac{PB}{PX}=\frac{QX}{QC} \Rightarrow \triangle BPX \sim \triangle XQC \Rightarrow \angle BXP= \angle XCQ \Rightarrow B,X$ and $C$ are collinear. Easy see that $\angle APQ = \angle BCA \Rightarrow BPQC$ is cyclical, If $Z$ is Miquel point of $\triangle ABC$ (Points $P,Q,X$) then $2\angle BAC=\angle BZC=\angle BOC$ and $\angle BXP=\angle BZP=\angle BCA=\angle QPA=\angle PQX =\angle PQZ + \angle ZCB$ Then $Z$ is tangency point of $\odot (BOC)$ and $\odot (PAQ)$ Why is $ \angle APQ=\angle BCA $ and how the last equality ( $ \angle BXP=\angle PQZ + \angle ZCB $ helps us to determinate that $ Z $ is the point of tangency?

navredras wrote: drmzjoseph wrote: Let $X$ be point such that $APXQ$ is an parallelogram $\Rightarrow \frac{PB}{PX}=\frac{QX}{QC} \Rightarrow \triangle BPX \sim \triangle XQC \Rightarrow \angle BXP= \angle XCQ \Rightarrow B,X$ and $C$ are collinear. Easy see that $\angle APQ = \angle BCA \Rightarrow BPQC$ is cyclical, If $Z$ is Miquel point of $\triangle ABC$ (Points $P,Q,X$) then $2\angle BAC=\angle BZC=\angle BOC$ and $\angle BXP=\angle BZP=\angle BCA=\angle QPA=\angle PQX =\angle PQZ + \angle ZCB$ Then $Z$ is tangency point of $\odot (BOC)$ and $\odot (PAQ)$ Why is $ \angle APQ=\angle BCA $ and how the last equality ( $ \angle BXP=\angle PQZ + \angle ZCB $ helps us to determinate that $ Z $ is the point of tangency? Hello, thanks for read my solution. 1. If $O_1$ is the circumcenter of $\triangle PAQ \Rightarrow \angle APQ=90^{\circ} - \angle O_1AQ= \angle BCA$ 2, $\angle PZB =\angle PQZ + \angle ZCB$ i.e. $\odot (PZQ)$ and $\odot (BZC)$ are tangents (If we drawing a line tangent to $\odot (PZQ)$ is easy with angles see that is tangent to $\odot (BZC)$ too)

Let $F$ be the point such that $\triangle FAB \sim \triangle FCA.$ Note that $$\frac{BP}{AP}=\frac{AQ}{CQ} \Longrightarrow F \in \odot (APQ).$$Let $AH$ be the $A$ altitude and $D$ be the antipode of $O$ in $\odot (BOC)$. Let $I, O'$ be the centres of $(I)$ and $\odot (BOC)$. Since $A, F, D$ are collinear and $AH \parallel OD$ we have $$\angle AFI=\angle FAH=\angle DAH=\angle ADO=\angle DFO'$$so $I, F, O'$ are collinear, establishing the tangency.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Le%20produit%20AB.AC.pdf p. 20... Sincerely Jean-Louis

Here's my solution

Now, the problem condition implies that this spiral similarity also sends $P$ to $Q.$ It follows that \[\angle APY = \angle YQC = 180 - \angle AQY\]thus $A, P, Q, Y$ are concyclic. It remains to show that $Y$ is the only common point between these two circles. Call the center of $(APQ)$ $X$ and the center of $(BOC)$ $K.$ We have \[\angle XYA = \angle XAY = \angle KTY = \angle KYT\]which implies that $X, Y, T$ are collinear, and the desired result follows.

Let $K$ be the midpoint of the symmedian chord, and $M, N$ be the midpoints of $AB, AC$. I claim $K \in (APQ)$. It suffices to show $K$ is the center of spiral similarity sending $PM \to QN$ and thus $PQ \to MN$. This is easy; clearly $AMKON$ is cyclic with diameter $AO$, so $\angle PMK = \angle QNK$. Furthermore,\[\frac{KM}{KN} = \frac{\delta(K, AB)}{\delta(K, AC)} = \frac{AB}{AC} = \frac{AB(\tfrac12 - t)}{AC(\tfrac12 - t)} = \frac{PM}{QN}\]where $t$ is $\tfrac{PB}{AB} = \tfrac{AQ}{AC}$ by the length condition, as desired. $K$ clearly lies on $(BOC)$, so the tangency point must be at $K$. Indeed, if we let $O_1$ be the center of the circle $I$, $O_2$ be the center of $(BOC)$, and $B'$ be the $B$ antipode in $(BOC)$, note that\[\angle O_1KA = \angle O_1AK = \angle BB'K = \angle B'KO_2\]where $AK$ clearly passes through $B'$ since $AK$ is a symmedian. Thus, $O_1, K, O_2$ are collinear so $I, (BOC)$ tangent at $K$. Edit: I am stupid. The spiral-sim part can be simplified. It is well known that $K$ is the center of spiral similarity sending $BA$ to $AC$, and this spiral-sim simultaneously sends $P \to Q$ and $M \to N$ due to ratios. This finishes.

Let $D$ be the center of spiral similarity sending $\overline{BA}$ to $\overline{AC}$, i.e. the so-called "$A$-dumpty point". Recall that $D$ lies on $(BOC)$ and that $\overline{AD}$ is a symmedian in $\triangle ABC$: both of these facts follow from angle chasing. Since the given condition is equivalent to $\frac{BP}{AP}=\frac{AQ}{QC}$, this spiral similarity also sends $P$ to $Q$. Therefore $D$ is also the intersection of $(APQ)$ and the circle passing through $B,P$ tangent to $\overline{PQ}$ (as well as the intersection of $(APQ)$ and the circle passing through $C,Q$ tangent to $\overline{PQ}$): in particular, $APQD$ is cyclic. I now claim that $D$ is the tangency point. Now, note that if the center of $I$ varies (not subject to the length condition) $\overline{PQ}$ is parallel to a fixed line, and by considering the case where the center is the midpoint of $\overline{AH}$ (where $H$ is the orthocenter as usual) we find that $\overline{PQ}$ is antiparallel to $\overline{BC}$ in general, i.e. $\triangle APQ \sim \triangle ACB$. Therefore, the tangent to $(APQ)$ at $A$ is parallel to $\overline{BC}$ If $T$ is the intersection of the tangents to $(ABC)$ at $B$ and $C$, then $T$ lies on $(BOC)$ and the tangent to $(BOC)$ at $T$ is parallel to $\overline{BC}$ as well. Hence we just need to prove the following. Claim: Let $\ell_1,\ell_2$ be (distinct) parallel lines. Let $A \in \ell_1,B \in \ell_2$, and $X$ lie on segment $\overline{AB}$. Let $\omega_1$ be tangent to $\ell_1$ at $A$ and pass through $X$, and $\omega_2$ be tangent to $\ell_2$ at $B$ and pass through $X$. Then in fact $\omega_1,\omega_2$ are tangent at $X$. Proof: Let $O_1,O_2$ be the centers of $\omega_1,\omega_2$ respectively. We have $\angle(\ell_1,\overline{AB})=\angle(\ell_2, \overline{AB})$, so $\angle AO_1X=\angle BO_2X$. Thus the homothety at $D$ sending $A$ to $B$ also sends $O_1$ to $O_2$, hence $X$ lies on $\overline{O_1O_2}$ which implies the desired result. $\blacksquare$

Let $E$ be the center of $I$, $F$ be the reflection of $A$ in $E$. $D$ be the foot of altitude from $A$ to $BC$, and $X$ be the circumcenter of triangle $BOC$ Let $AE = x \implies AF = 2x \implies AP = 2xsinB$ Now $4x^2 sinBsinC = (b-2xsinC)(c-2xsinB) = 4(RsinB - xsinC)(RsinC-xsinB)$ $\implies x= R\frac{sinB sinC}{sin^2 B + sin^2 C}$ Also $OX = \frac{OB}{2sin(90-A)} = \frac{R}{2cosA} $ Let $M$ and $N$ be the feet of perpendicular from $O$ to $BC$ and $AD$ and $L$ be the foot of perpendicular from $X$ to $AD$ Now $OM = RcosA \implies EL = AD+DL-AE = AD-AE+MX = 2RsinBsinC - \frac{Rcos2A}{2cosA} - x$ Also $LX = MD = 2RsinBcosC - RsinA = Rsin(B-C)$ Now we know $EX^2 = EL^2 + LX^2$ simplifying which we find $EX = x+ \frac{R}{2cosA}$ which is the sum of the 2 radii Hence proved

After inversion at $A$ it suffices to prove that for $\triangle ABC$ if $P,Q$ are points on $AB,AC$ with $\frac{AC}{AQ}+\frac{AB}{AP}=1$ and $PQ \perp AO$ then $PQ$ is tangent to the reflection of $(ABC)$ across $BC.$ We claim that the desired tangency point is $A',$ the reflection of $A$ across the midpoint $M$ of $BC.$ It is clear that this point is on the reflection of $(ABC)$ across $BC.$ Furthermore, if this circle has center $O'$ then we see $O'A' \parallel AO \perp PQ,$ so it suffices to show that $P,A',Q$ collinear. To do this, consider the midpoints $P',Q'$ of $AP',AQ'.$ It suffices to show that $P',M,Q'$ collinear. We have $\frac{AC}{AQ'}+\frac{AB}{AP'}=2,$ so $2AP'\cdot AQ'=AB\cdot AQ'+AC\cdot AP'.$ This rearranges to $AP'\cdot AQ'-AP'\cdot AC=-AP'\cdot AQ'+AB\cdot AQ',$ or $\frac{AP'}{AQ'}=-\frac{AP'-AB}{AQ'-AC},$ or $\frac{AP'}{P'B}\cdot \frac{CQ'}{Q'A}=-1.$ Thus from $BM=MC$ we finish by Menelaus.

Solved with dolphinday Let $D_O$ the A-dumpty point, note that $D_O$ is basically the Isogonal conjugate of the A-humpty point, from there you can derive basic propeties of it such as $D_O$ lies in $(BOC)$ and $(AO)$, and also that it is center of spiral sim sending $AB \to CA$, hence we notice that by the ratios given it sends $BA \to PQ \to AC$, hence by degenerate complete quad miquel propeties (or just angle chase) we see that $D_O$ lies in $I$, let $O'$ the center of $I$ and $O_1$ the center of $(BOC)$, let $AD_O \cap (BOC)=T$, note that $AD_O$ is symedian thus $T$ is the point where the tangents from $B,C$ to $(ABC)$ meet at, let $D_OO \cap AO'=X$, then $X$ also lies on $I$ since $\angle AD_OO=90$, Note that $O'$ is midpoint of $AX$ and $O_1$ is midpoint of $OT$, since triangles $\triangle AD_OX$ and $\triangle TD_OO$ share the same 90 degree angle we have that $O', D_O, O_1$ are colinear thus $I$ is tangent to $(BOC)$ at $D_O$, thus we are done

It is well-known that the given product condition implies that there exists a point $E$ on $\overline{BC}$ with $APEQ$ a parallelogram. (To see this, just note that for this point $E$, the triangles $BPE$ and $EQC$ are similar.) Now, I claim that $(APQ)$, $(BPE)$, $(CQE)$, $(BOC)$ concur at a point $X$ on $\overline{AF}$. To see this, let $X = (BFE) \cap (CQE)$ and note that $X$ lies on $(APQ)$ by Miquel point and $\angle BQF + \angle CPF = \angle BOC$, so $X$ lies on $(BOC)$. Next, notice that as the circumcenter $O_1$ of $(APQ)$ lies on the $A$-altitude, it follows by isogonal conjugates that triangles $APQ$ and $ACB$ are similar, so $BPQC$ is cyclic. Then $A$ is the radical center of $(BEP)$, $(CEQ)$, and $(BPQC)$, so $A$ lies on $\overline{XF}$. Finally, notice that $$\angle AXC = \angle C = \angle APE = \angle APX + \angle XPE = \angle XBC + \angle XAQ,$$so the tangent to $(BOC)$ at $X$ coincides with the tangent to $(APQ)$ at $X$. It follows that the two circles are tangent to each other.

Start by constructing $R$ such that $APRQ$ is a parallelogram: The length condition along with $\angle RPB = \angle CQR = \angle A$ gives us $\triangle RPB \sim \triangle CQR$, which can be further be used to show that $R$ lies on $BC$, with \[\angle QRC + \angle PRQ + \angle BRP = \angle B + \angle A + \angle C = 180.\] Consider the Miquel point $M$ of $\triangle ABC$ wrt $PQR$. Notice $M \in (BOC)$, as \[\angle BMC = \angle BPR + \angle CQR = 2 \angle A = \angle BOC.\] We also have $M \in AR$, as \[\angle AMP = \angle AQP = \angle B = \angle RMP.\] At this point, we deduce $AP$ is a symmedian as it bisects the antiparallel $PQ$, so it passes through the antipode of $O$ on $(BOC)$, and thus the two desired circles are tangent from homothety. $\blacksquare$

Let $\omega$ be the circumcircle of $\triangle BOC$, and let $\Gamma$ be the circumcircle of $\triangle APQ$. Let $M$ be the midpoint of arc $\widehat{BC}$. Then we make the following claim: Claim: $\triangle XBA \sim \triangle XAC$, and as $BP/PA = AQ/QC$, we must have $\triangle XDP = \triangle XAQ$, and $\triangle XPA \sim \triangle XQC$. Proof: Let $\angle XBO = \alpha$. Observe $XM$ bisects $\angle BXC$, so that $\angle BXM = \angle A$, and so $\angle BXA = \angle AXC$. Now \begin{align*} \angle XBA &= \angle B - \angle XBC \\ &= \angle B - (\alpha + \angle OBC) \\ &= \angle B - (\alpha + 90^{\circ} - \angle A) \\ &= 180^{\circ} - \angle C -\alpha - 90^{\circ} \\ &= 90^{\circ} - \angle C - \alpha. \end{align*}Similarly, \begin{align*} \angle XAC &= 180^{\circ} - (\angle AXC + \angle XCA) \\ &= 180^{\circ} - (180^{\circ} - \angle A + \alpha + 90^{\circ} - \angle B) \\ &= \angle A - \alpha - 90^{\circ} + \angle B \\ &= 90^{\circ} - \angle C - \alpha, \end{align*}so that $\triangle XBA \sim \triangle XAC$. $\square$ We now make another claim: Claim: $X$ lies on $\Gamma$. Proof: Observe that \[\angle PXA = 180^{\circ} - \angle PAX - \angle XPA, \text{ and } \angle QXA = 180^{\circ} - \angle XQA - \angle QAX = \angle XQC - \angle QAX.\]Hence, \begin{align*} \angle PXQ &= \angle PXA + \angle QXA \\ &= (180^{\circ} - \angle PAX - \angle XPA) + (\angle XQC - \angle QAX) \\ &= 180^{\circ} - \angle A, \end{align*}as desired. $\square$ We now finally show that the circumcenters of $\Gamma, \omega$ ($O_{\Gamma}$ and $O$ respectively), are collinear with $X$: $\angle MXO = \angle XMO$, and as the altitude from $A$ and line $OM$ are parallel, and further as $A, X, M$ are by definition collinear, then \begin{align*} \angle O_{\Gamma}XA &= \angle O_{\Gamma}AX \\ &= \angle AMO \\ &= \angle XMO \\ &= MXO, \end{align*}so that $O_{\Gamma}, X, O$ are collinear as $A, X, M$, as desired. $\blacksquare$

A really beautiful problem. Once you figure out how to deal with the really weird angle condition, the rest is standard. We start off with the following observation. Claim : Points $B,C,P$ and $Q$ lie on the same circle. Proof : Simply notice that $$\angle PQC = \angle PQK + \angle KQC = \angle PAK + 90 = 90 +\angle BAD = 180 - \angle DBA = 180 - \angle PBC$$which implies our claim. Then, let $F$ be the point such that $AQFP$ is a parallelogram. Let $L= (BPF) \cap (FQC)$. Then, we have the following key claim. Claim : Point $F$ in fact lies on the line $\overline{BC}$. Proof : Let $F'$ be the point on $BC$ such that $PF' \parallel AB$. Then, notice that since $\triangle BPF' \sim \triangle BAC$, $$\frac{BP}{PF'}=\frac{AB}{AC}$$which gives us that $$\frac{AC}{AQ}= 1+ \frac{CQ}{AQ}=1+\frac{AP}{BP}=\frac{AB}{BP}=\frac{AC}{PF'}$$Thus, we require $AF'=AQ$ and thus $APF'Q$ is indeed a parallelogram. Now, consider the following. \begin{align*} \angle BCF &= \angle BPF = \angle{A}\\ \angle CLF &= \angle CQF = \angle A\\ \angle BLC &= 2\angle A = \angle BOC \end{align*}which means that $L$ must lie on $(BOC)$. Also, \begin{align*} \angle LPA &= 180 - \angle LPB\\ &= \angle LFB\\ &= 180 - \angle LFC\\ &= \angle LQC \end{align*}and thus $L$ lies on $(APQ)$ as well. But then, \begin{align*} \angle ALP &= \angle AQP = \angle B\\ \angle PLF &= 180 - \angle B \end{align*}So clearly $L$ also lies on $AF$. Let $T$ lie on the line $\ell$ such that $\ell$ is tangent to $(APQ)$ at $L$. Then, $\angle TLQ = \angle LAQ$ which in turn gives \begin{align*} \angle CLT &= \angle CLQ - \angle TLQ\\ &= \angle QFC - (180 - \angle AFC - \angle C)\\ &= \angle B + \angle C - \angle LFB\\ &= 180 - \angle A - \angle LFB\\ &= \angle BPL - \angle A\\ &= \angle FPL\\ &= \angle FBL\\ &= \angle CBL \end{align*}Therefore, $\ell$ is also tangent to $(BOC)$ at $L$. Thus, indeed $\omega$ is tangent to the circumcircle of $\triangle BOC$ as required.

Here is my solution: Let $D$ be the feet of the altitude from $A$ to $BC$, and let $A'$ be a point on $AD$ such that $AA'$ is the diameter of $\omega$ , $\odot I=\omega$ $\textbf{Claim:}$ Points $B,P,Q$ and $C$ are concyclic. $\textbf{Proof:}$

Let $Q_A$ be t the $A$-dumpty point of the triangle $\triangle ABC$ $\textbf{Claim:}$ $Q_A \in \omega$ $\textbf{Proof:}$

$\textbf{Claim:}$ $Q_A \in \odot ( BOC)$ $\textbf{Proof:}$

$\textbf{Claim:}$ $\omega$ is tangent to the circumcircle of triangle $\triangle BOC$ $\textbf{Proof:}$

By the ratio condition, we get that the lines parallel to $AB,AC$ through $P,Q$ respectively intersect on $BC$ can at $R$. Let $X = \omega \cap (BPR)$ by miquel's theorem, $X \in (CQR)$ \[ \angle BXC = \angle BXR + \angle CXR = \angle BPR + \angle CQR = 2A \]Also, $\angle BOC = 2A$; hence, $B,X,O,C$ are concyclic. Next, observe that \[ \angle AXP = \angle AXQ = B = 180 - \angle RXP \]Hence $X \in AR$ Finally as $APRQ$ is a parallelogram and $AR$ bisects $PQ$ which is antiparallel to $BC$ we get $AR$ is the $A$ symmedian. Hence it goes through the antipode of $O$ wrt $(BOC)$ and hence we are done by by homothety.

Let $D$ be the $A$-dumpty point in $\triangle ABC$. We claim this is the desired tangency point. The length condition implies that there exists a point $R$ on $\overline{BC}$ such that $APRQ$ is a parallelogram. This in turn implies that the midpoint of $\overline{PQ}$ lies on $\overline{MN}$, where $M$ and $N$ are the midpoints of $\overline{AB}$ and $\overline{AC}$. Claim: The Miquel point of the concave quadrilateral $MPQN$ is $D$. Proof: Since $D$ is the center of spiral similarity from $\overline{AB}$ to $\overline{CA}$, we just need to show that $AMPB \sim CNQA$. We just need to show $\tfrac{MP}{AB} = \tfrac{QN}{AC}$, which combined with the given length condition, implies the similarity. But this is just Menealus on tranversal $MN$ WRT $\triangle APQ$. So, $APQD$ is cyclic. Let $T$ be the intersection of the tangents to $(ABC)$ at $B$ and $C$. Then, $A$, $D$ and $T$ are collinear. The fact that the center of $(APQ)$ lies on the $A$-altitude implies that $A$ is at the ``top" of $(APQ)$. Meanwhile, $T$ is at the bottom of $(BOC)$, so the tangency follows.