Let $H$ be the orthocenter of $\triangle BIC$,$S$ the midpoint of $IK$ and $S'$ its invers WRT the incircle,let $R,Q$ be the touch points of incircle with $AB,AC$,$W$ the reflection of $I$ wrt $BC$ Lemma(well-known):The pole of $LM$ is $H$ The main idea will be to prove that polar of $P$ is perpendicular to $RQ$. Point $P$ lies on the polar of $H$ and hence its polar passes thru $H$ ,polar of $S'$ passes thru $P$and hence the polar of $P$ is $HS'$.The inversion with wrt $\odot I$ takes the midpoint of $ID$ to $W$ and hence $K$ is the midpoint of $IS'$.By the last conclusion $S'$ is the reflection of $I$ over $K$ and so as $I$ is the orthocenter of $HBC$ we have that $HS'$ contains the center of $\odot HBC$.Its well known that the feet of perpendicular form $C$ to $IB$,$B$ to $IC$ lie on $PQ$ and let those be $R',Q'$.$P'Q'$and $BC$ are anti-parallel and hence as $HS',HI$ are isogonals we have $HS'\perp R'Q'\equiv RQ$.$\clubsuit$

Denote by $\omega$ the circumcircle of $\triangle BIC$ and let $\gamma$ be the circle with center $K$ passing through $D.$ Let $X$ be the midpoint of arc $\widehat{BC}$ on $\odot(ABC)$ and let $T$ be the midpoint of $\overline{ID}.$ By the Incenter/Excenter Lemma, $X$ is the center of $\omega.$ Claim: $LM$ is the radical axis of $\omega$ and $\gamma.$ Proof: Since $AX$ is the bisector of $\angle BAC$, the reflection $B^*$ of $B$ in $AX$ is the second intersection of $AC$ with $\odot(BIC).$ Since $AB^* = AB$, we have \[ \text{pow}(L, \omega) = LC \cdot LB^* = \left(\frac{b}{2}\right)\left(c - \frac{b}{2}\right) = \left(\frac{c}{2}\right)^2 - \left(\frac{b - c}{2}\right)^2 = KL^2 - KD^2 = \text{pow}(L, \gamma). \]Thus, $L$ lies on the radical axis of $\omega$ and $\gamma.$ Similarly $M$ lies on said radical axis, and the claim follows. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 969.6063173076674, xmax = 1067.8477329910697, ymin = 987.7286905570053, ymax = 1037.655389958556; /* image dimensions */ pen evevev = rgb(0.8980392156862745,0.8980392156862745,0.8980392156862745); pen evefev = rgb(0.8980392156862745,0.9372549019607843,0.8980392156862745); pen qqwuqq = rgb(0.,0.39215686274509803,0.); filldraw((998.8132125710364,1032.3295768403102)--(992.8955461776062,1007.2387857490346)--(1021.2800266331659,1007.6574023678552)--cycle, evevev); filldraw((1005.1944554804243,1011.3426291473904)--(1006.0487248750474,1011.7579328039642)--(1005.6334212184736,1012.6122021985873)--(1004.7791518238505,1012.1968985420135)--cycle, evefev, qqwuqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((998.8132125710364,1032.3295768403102)--(992.8955461776062,1007.2387857490346)); draw((992.8955461776062,1007.2387857490346)--(1021.2800266331659,1007.6574023678552)); draw((1021.2800266331659,1007.6574023678552)--(998.8132125710364,1032.3295768403102)); draw(circle((1007.087786405386,1007.4480940584449), 3.7947839968939716)); draw(circle((1007.1981237584237,999.9666222295392), 16.045188927201252)); draw((1003.173854993174,1015.498954591893)--(1003.2934150351269,1007.3921343618717), dotted); draw((995.8543793743213,1019.7841812946724)--(1021.1534905655748,1020.1572946374557), red); draw((1003.173854993174,1015.498954591893)--(1007.087786405386,1007.4480940584449)); draw((1003.2336350141504,1011.4455444768823)--(1021.1534905655748,1020.1572946374557), red); draw(circle((1006.94442771478,1017.1685919285534), 17.203840362208762)); draw((998.8132125710364,1032.3295768403102)--(1007.1981237584237,999.9666222295392), blue); /* dots and labels */ dot((998.8132125710364,1032.3295768403102),linewidth(3.pt) + dotstyle); label("$A$", (998.1742403733058,1032.9089952172428), NW * labelscalefactor); dot((992.8955461776062,1007.2387857490346),linewidth(3.pt) + dotstyle); label("$B$", (991.4576440412592,1006.6247149044981), W * labelscalefactor); dot((1021.2800266331659,1007.6574023678552),linewidth(3.pt) + dotstyle); label("$C$", (1021.9509913887509,1007.1620426110618), E * labelscalefactor); dot((1003.2934150351269,1007.3921343618717),linewidth(3.pt) + dotstyle); label("$D$", (1002.2041981725338,1005.8635006535327), S * labelscalefactor); dot((1003.173854993174,1015.498954591893),linewidth(3.pt) + dotstyle); label("$I$", (1003.368408203422,1015.7592859160823), NE * labelscalefactor); dot((1007.087786405386,1007.4480940584449),linewidth(3.pt) + dotstyle); label("$K$", (1007.2192567671286,1005.8187233446524), E * labelscalefactor); dot((1010.0466196021011,1019.9934896040827),linewidth(3.pt) + dotstyle); label("$L$", (1010.2193364621095,1020.281794112994), NE * labelscalefactor); dot((995.8543793743213,1019.7841812946724),linewidth(3.pt) + dotstyle); label("$M$", (994.3233918095991,1019.9683529508318), N * labelscalefactor); dot((1003.2336350141504,1011.4455444768823),linewidth(3.pt) + dotstyle); label("$T$", (1001.6220931570898,1011.2367777191705), N * labelscalefactor); dot((1021.1534905655748,1020.1572946374557),linewidth(3.pt) + dotstyle); label("$P$", (1021.3241090644266,1020.416126039635), NE * labelscalefactor); label("$\gamma$", (1010.7416852690592,1009.4456853639579), NE * labelscalefactor); label("$\omega$", (997.1593411686727,1014.7144289122208), NE * labelscalefactor); dot((1007.1981237584237,999.9666222295392),linewidth(3.pt) + dotstyle); label("$X$", (1006.3684878984027,998.2961354527596), E * labelscalefactor); dot((1016.1700546259426,1013.2689733081288),linewidth(3.pt) + dotstyle); label("$B_*$", (1016.3538277787121,1013.5204204720665), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Back to the proof at hand, notice that $T$ lies on the radical axis of $\gamma$ and the degenerate $\odot(I).$ Moreover, $TP \perp KI$, so $TP$ is the radical axis of $\gamma$ and $\odot(I).$ By the claim, $P$ is the radical center of $\omega, \gamma, \odot(I)$, so $P$ lies on the radical axis of $\omega$ and $\odot(I).$ Hence $PI$ is tangent to $\omega$, which gives $PI \perp AI$, as desired. $\square$

Let $\omega$ be the incircle, $H$ be the orthocenter of $\triangle BIC$, and $G$ be the antipode of $H$ in $\odot(BHC)$. It is easy to see that $G$ lies on the polar of $E$ with respect to $\omega$, so by La Hire's Theorem, $P$ lies on the polar of $G$ with respect to $\omega$ as well. It is well-known that $LM$ is the polar of $H$ with respect to $\omega$, so it follows that line $GH$ is the polar of $P$ with respect to $\omega$. Finally, since line $HG$ is the reflection of line $AI$ about $K$, it follows that they are parallel, so we're done.

the incircle touch $AB$ and $AC$ at $F$ and $E$ rep, $BI \cap ED \equiv X, CI \cap FD \equiv Y, XY \equiv LM$ (well-known), $BI \cap FD \equiv X', CI \cap \equiv Y'$ is clear that $X'Y' \perp AI$. Denote $R$ the midpoint of $ID$ and $Q \equiv IR \cap XY$ so $Q$ is midpoint of $YX \Rightarrow QR \perp X'Y'$ also $I$ is orthocenter of $\triangle QRP \Rightarrow IP \perp QR \Rightarrow X'Y' \parallel IP \Rightarrow AI \perp IP$

Let $E$ and $F$ the points of tangency of the incircle with $AC$ and $AB$, let $N$ the midpoint of $ID$. Working in complex numbers where the incircle of $\triangle ABC$ is the unit circle $\Longrightarrow$ $i=0$. Suppose that $d=1$ $\Longrightarrow$ $n=\frac{1}{2}$. Finding $a$, $b$​​, $c$, $m$, $k$, $l$ in funtion of $e$ and $f$ we get: $a=\frac{2ef}{e+f},$ $b=\frac{2f}{1+f},$ $c=\frac{2e}{1+e},$ $m=\frac{f}{1+f}+\frac{ef}{e+f},$ $l=\frac{e}{1+e}+\frac{ef}{e+f},$ $k=\frac{e}{1+e}+\frac{f}{1+f}$. From $PN\perp KI$ we get $\frac{p-n}{\overline{p}-\overline{n}}=-\frac{k-i}{\overline{k}-\overline{i}}$ $\Longrightarrow$ $\frac{p-\frac{1}{2}}{\overline{p}-\frac{1}{2}}$ $=$ $-\frac{k}{\overline{k}}=-\frac{e+f+2ef}{2+e+f}...(1)$ From $M$, $L$, $P$ are collinear we get: $\frac{p-l}{\overline{p}-\overline{l}}=\frac{m-l}{\overline{m}-\overline{l}}=\frac{\frac{f}{1+f}-\frac{e}{1+e}}{\frac{1}{1+f}-\frac{1}{1+e}}=-1$ $\Longrightarrow$ $p-l=\overline{l}-\overline{p}...(2)$ By $(1)$ and $(2)$ we get: $p=\frac{ef(e+1)(f+1)}{(e+f)(ef-1)}$ and $\overline{p}=-\frac{(e+1)(f+1)}{(e+f)(ef-1)}$ $\Longrightarrow$ $AI\perp PI$ $\Longleftrightarrow$ $\frac{a-i}{\overline{a}-\overline{i}}=-\frac{p-i}{\overline{p}-\overline{i}}$ $\Longleftrightarrow$ $\frac{a}{\overline{a}}=-\frac{p}{\overline{p}}=ef$ $\Longleftrightarrow$ $\frac{a}{\overline{a}}=\frac{\frac{2ef}{e+f}}{\overline{\frac{2ef}{e+f}}}=ef$ which it is true. Hence $\measuredangle PIA=90^{\circ}$.

Let's recall the classic "midpoints of altitudes lemma" and its configuration: $S$ is the midpoint of the $A$-altitude, and $E,F$ are the topmost points of the $A$-excircle and incircle. It's known that $K$ is the midpoint of $\overline{DE}$. By homothety, $A,F,E$ are collinear, and so after an affinity from line $BC$ with ratio $1/2$, we get the line of $S,I,E$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7.7893029041155195cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.0057065599480834, xmax = 18.249444594667686, ymin = -4.14653182629387, ymax = 6.358037964415124; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); /* draw figures */ draw((0.8398940024876984,4.056377697258126)--(0.8290125451759012,1.8347159228505114), dotted); draw((0.8398940024876984,4.056377697258126)--(3.1037120882181446,-0.39814037827332144), qqwuqq); draw((0.8290125451759012,1.8347159228505114)--(3.1037120882181446,-0.39814037827332144), qqwuqq); draw((1.5220941067768126,1.1543836403972414)--(2.309120112487158,-0.3942485533502777), linetype("2 2")); draw(circle((2.5914120762834707,2.9369414172678483), 2.0786902481590963), linewidth(0.4) + red); draw(circle((1.5220941067768126,1.1543836403972414), 1.5447588973617619), linewidth(0.4)); draw((0.8398940024876984,4.056377697258126)--(-0.6,-0.38)); draw((-0.6,-0.38)--(5.218240224974315,-0.4084971067005554)); draw((5.218240224974315,-0.4084971067005554)--(0.8398940024876984,4.056377697258126)); draw((0.11994700124384922,1.8381888486290632)--(4.342930150079244,1.817505137277571)); draw((4.342930150079244,1.817505137277571)--(1.5183111217664917,0.3820134559850037)); draw((1.5296600767974537,2.6991240092217157)--(1.514528136756171,-0.39035672842723396)); /* dots and labels */ dot((0.8398940024876984,4.056377697258126),dotstyle); label("$A$", (0.9091168676653525,4.2294348602022644), NE * labelscalefactor); dot((-0.6,-0.38),dotstyle); label("$B$", (-0.5272575847709685,-0.20082851116759257), NE * labelscalefactor); dot((5.218240224974315,-0.4084971067005554),dotstyle); label("$C$", (5.287463090151969,-0.23543994375641958), NE * labelscalefactor); dot((0.11994700124384922,1.8381888486290632),linewidth(3.pt) + dotstyle); label("$M$", (0.1822767832999852,1.9450803093396818), NE * labelscalefactor); dot((3.029067113731007,1.8239402952787853),linewidth(3.pt) + dotstyle); label("$L$", (3.106942837055868,1.9277745930452683), NE * labelscalefactor); dot((1.5220941067768126,1.1543836403972414),linewidth(3.pt) + dotstyle); label("$I$", (1.5840398031474792,1.2528516575631417), NE * labelscalefactor); dot((2.309120112487158,-0.3942485533502777),linewidth(3.pt) + dotstyle); label("$K$", (2.3801027526905005,-0.2873570926396601), NE * labelscalefactor); dot((1.514528136756171,-0.39035672842723396),linewidth(3.pt) + dotstyle); label("$D$", (1.5840398031474792,-0.2873570926396601), NE * labelscalefactor); dot((3.1037120882181446,-0.39814037827332144),linewidth(3.pt) + dotstyle); label("$E$", (3.1761657022335217,-0.2873570926396601), NE * labelscalefactor); dot((1.5296600767974537,2.6991240092217157),linewidth(3.pt) + dotstyle); label("$F$", (1.6013455194418926,2.810366124060357), NE * labelscalefactor); dot((1.5183111217664917,0.3820134559850037),linewidth(3.pt) + dotstyle); label("$T$", (1.5840398031474792,0.49140014060894754), NE * labelscalefactor); dot((0.8290125451759012,1.8347159228505114),linewidth(3.pt) + dotstyle); label("$S$", (0.8918111513709389,1.9450803093396818), NE * labelscalefactor); dot((4.342930150079244,1.817505137277571),linewidth(3.pt) + dotstyle); label("$P$", (4.404871559136881,1.9277745930452683), NE * labelscalefactor); dot((2.463193295160249,0.8622093642164187),linewidth(3.pt) + dotstyle); label("$U$", (2.535854199340222,0.9586544805581121), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $T$ be the midpoint of $\overline{ID}$. Since $\overline{IK}$ and $\overline{FE}$ are parallel, $TP$ hits $AE$ perpendicularly at $U$. Clearly, $ASUP$ is cyclic with diameter $\overline{AP}$. We need only show that $I$ also lies on its circle. (So, the crux of this proof is introducing new points to connect that pesky $P$ to the rest of the diagram, and then forgetting $P$.) But $ASIU$ being cyclic is equivalent by PoP to $EI\cdot ES=EU\cdot EA$, but as $AS\parallel FI$, we just want $EI^2=EU\cdot EF$. This is but a simple computation: it's easy to check that if $DE=a$, $DF=b$, $EF=c$, then both sides are equal to $c\left(\frac{a^2}{c}+\frac{b^2}{4c}\right)$, so we are done. $\blacksquare$

A rather straight forward solution with not much insight (and computation). Assume for the sake of convenience, $AB<AC$. We shall exploit the perpendicularity lemma. Let $T$ be the midpoint of $ID$. Let the line through $T$ perpendicular to $IK$ meet the $A$ midline at $P$. We show that $\angle PIA=90^{\circ}$. Clearly, we have $KP^2-PI^2=KT^2-IT^2=TD^2+DK^2-IT^2=DK^2$. It suffices to show that $PI^2=AP^2-AI^2$ which is equivalent to saying that $KP^2-AP^2=DK^2-AI^2$. Let $Z$ be the midpoint of arc $BC$ of the circumcircle and let $ZK$ meet $LM$ at $W$. Clearly, $KP^2-PZ^2=WK^2-WZ^2$ and so it suffices to showing that $AP^2-PZ^2=AI^2-DK^2+WK^2-WZ^2$. Therefore, it suffices to showing that $AI^2-IZ^2=AI^2-DK^2+WK^2-WZ^2$. This reduces to $DK^2=IZ^2+WK^2-WZ^2$. Let $ZK$ meet the line through $A$ parallel to $BC$ at $X$ and the circumcircle of triangle $ABC$ at $Y$. By the "shooting lemma", $ZI^2=ZK\cdot ZY$ and $WK^2-WZ^2=-ZM\cdot ZX$. We get that the requested condition is equivalent to $DK^2=ZK\cdot XY$. Now comes the part with calculations. Let $BC=a,CA=b,AB=c$ and the usual notations. We get that $XY=\frac{b^2-c^2}{2a}\cdot \tan \left(\frac{B-C}{2}\right)$ and $ZK=\frac{a}{2}\cdot \tan \frac{A}{2}$ and $DK=\frac{b-c}{2}$. Thus, we only need to show that $\tan (A/2)\cdot \tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}$. This is rather easy, considering that the RHS evaluates to \begin{align*} \frac{b-c}{b+c}=\frac{\sin B-\sin C}{\sin B+\sin C}=\frac{2\sin (\frac{B-C}{2})\cos (\frac{B+C}{2})}{2\sin (\frac{B+C}{2})\cos (\frac{B-C}{2})}=\tan (A/2)\cdot \tan \left(\frac{B-C}{2}\right) \end{align*}which holds because of the standard sum-to-product trig formulas. The conclusion holds.

This problem was proposed by Burii. Let me post yet another solution: Let $X$, $Y$, $Z$ be reflections of $D$ about $I$, $K$, $P$ respectively. [asy][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(13.4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 9.1, ymin = -2.86, ymax = 6.3; /* image dimensions */ /* draw figures */ draw(circle((-0.1,0.49), 2.53)); draw((-1.88,5.48)--(-3.06,-2.04)); draw((-3.06,-2.04)--(6.52,-2.04)); draw((6.52,-2.04)--(-1.88,5.48)); draw((-0.1,0.49)--(1.73,-2.04)); draw((3.56,-2.04)--(-1.88,5.48)); draw((-1.88,5.48)--(6.79,5.48)); draw((3.35,1.72)--(-0.1,0.49)); draw((-0.1,0.49)--(-1.88,5.48)); draw((-0.1,3.02)--(6.79,5.48)); draw((-0.1,-2.04)--(-0.1,5.48)); draw((-0.1,0.49)--(6.79,5.48)); draw((-0.1,-2.04)--(6.79,5.48)); draw((-0.1,-0.77)--(3.35,1.72)); draw((-2.47,1.72)--(3.35,1.72)); /* dots and labels */ dot((-1.88,5.48),dotstyle); label("$A$", (-1.8,5.6), NE * labelscalefactor); dot((-3.06,-2.04),dotstyle); label("$B$", (-3.36,-1.9), NE * labelscalefactor); dot((6.52,-2.04),dotstyle); label("$C$", (6.6,-1.92), NE * labelscalefactor); dot((-0.1,0.49),dotstyle); label("$I$", (-0.36,0.44), NE * labelscalefactor); dot((-0.1,-2.04),dotstyle); label("$D$", (-0.28,-2.4), NE * labelscalefactor); dot((-2.47,1.72),dotstyle); label("$M$", (-2.8,1.8), NE * labelscalefactor); dot((2.32,1.72),dotstyle); label("$L$", (2.4,1.84), NE * labelscalefactor); dot((1.73,-2.04),dotstyle); label("$K$", (1.82,-1.92), NE * labelscalefactor); dot((3.56,-2.04),dotstyle); label("$Y$", (3.64,-1.92), NE * labelscalefactor); dot((1.68,-9.56),dotstyle); label("$A'$", (-4.3,6.3), NE * labelscalefactor); dot((-0.1,3.02),dotstyle); label("$X$", (-0.02,3.14), NE * labelscalefactor); dot((3.35,1.72),dotstyle); label("$P$", (3.48,1.42), NE * labelscalefactor); dot((6.79,5.48),dotstyle); label("$Z$", (6.88,5.6), NE * labelscalefactor); dot((-0.1,5.48),dotstyle); label("$E$", (-0.02,5.6), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We shall prove that $A, X, Y$ are collinear. Since $K$ is the midpoint of segments $BC$, $DY$, we have $BD = CY$. Therefore $Y$ is the common point of segment $BC$ and the $A$-excircle. Consider homothety centered at $A$ mapping incircle to $A$-excircle. It's easy to see that this homothety maps $X$ to $Y$. Therefore $A,X,Y$ are collinear. Consider homothety with centre $D$ and ratio $2$. We easily see that $AZ \parallel LM$. Since $IX \perp BC \parallel LM$, we have $$IX \perp AZ.$$ Moreover $IZ \parallel g \perp KI \parallel XY$, where $g$ is the line perpendicular to $IK$ passing through the midpoint of $ID$. Since $A,X,Y$ are collinear, we have $$IZ \perp AX.$$ Thus $X$ is the orthocentre of triangle $AIZ$. Therefore $ZX \perp AI$. Since $ZX \parallel PI$, we conclude that $PI \perp AI.$ In other words, $$\angle PIA = 90^\circ.$$

Let H be the orthocenter of BIC and K' be the reflection of I in K. Clearly by la Hire we get that HK' is the polar of P wrt the incircle. It suffices to show that HK' || AI. Note that I is the orthocenter of BIC and thus K' is the antipode of H in (BHC). By a straightforward angle chase the angle bw HK' and BC equals 90°+C/2-B/2 which is just that between AI and BC and we're done.

My solution Let $N,J$ be the midpoint of $AD,ID$. It's well-known that $\overline{N,I,K}$ (Newton's theorem in triangle). It's easy to see $I$ is the orthocenter of $PNJ$. So $IP \perp NJ$ and then $IP \perp IA$

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设 ${DI}$ 交 $\triangle{ABC}$ 于另一点 ${T},$ $AT,BC$ 相交于点 $S.$ 由内切圆性质知 ${S}$ 为 $A-$ 旁切圆切点且 $KD=KS,$ 从而 $KI\| AS.$ 设 $AS,NP$ 相交于点 $V,$ 则 $\angle AVP=\frac{\pi}2.$ 设过 ${A}$ 垂直于 $BC$ 垂线分别交 $LM,BC$ 于点 $Q,E,$ 则 ${Q}$ 为 ${AE}$ 中点$.$ 由 $TD\| AE,$ ${I}$ 为 $TD$ 中点$,$ ${Q}$ 为 $AE$ 中点$,$ 有 $S,I,Q$ 三点共线$.$ 由 $\angle AQP=\angle AVP=\frac{\pi}2,$ $A,Q,V,P$ 四点共圆$.$ 由 $SV\cdot ST=TS^2-TV\cdot TS=(TD^2+DS^2)-TN\cdot TD=DS^2+DN\cdot DT=DS^2+DI^2=IS^2,$ 可以得到 $\triangle SIV\sim\triangle STI.$ 从而 $\angle SIV=\angle STI=\angle VAQ,$ 即 $A,Q,I,V$ 四点共圆$,$ 故 $A,Q,I,V,P$ 五点共圆$,$ 进而 $\angle AIP=\angle AQP=\frac{\pi}2.\blacksquare$

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