Just some length bashing and showing that $I$ is the othocenter of triangle $APQ$

tarzanjunior wrote: Just some length bashing and showing that $I$ is the othocenter of triangle $APQ$ That's really cool.

Dear Mathlinkers, see also http://www.artofproblemsolving.com/community/c6t48f6h1085209_angle_bisector_perpendicular_to_line_with_2_incenters http://jl.ayme.pagesperso-orange.fr/Docs/Le%20theoreme%20de%20Feuerbach-Ayme.pdf p. 17-19. Sincerely Jean-Louis

Show $BI \perp AQ$ and $CI \perp AP$ giving $I$ is orthocenter of $\Delta APQ$. For part 2, just chase lengths $AI,PI,QI,PQ$ by law of sines on $\Delta AIP, \Delta AIQ$ and $\Delta PIQ$.

Here's another solution. In ABC, B, P, I are co-linear and C, Q, I are co-linear. Find angle PIQ (which is 135). From angle bisectors angle PAQ=45. Reflect PIQ about PQ to get a new point K. We find that quad. APKQ is cyclic. Reflection of point I i.e K lies on circumcircle of APQ forcing the point to be the orthocenter of APQ. For the second part some length chasing using sine rule will give the desired result.

There is another simpler but longer solution using coordinate geometry, setting D as the origin and AD as Y axis and BC as X axis.

For the first part just angle chase to show $BI$ is perpendicular to AQ and similarly $CI$ perpendicular to $AP$ . So $I$ is the orthocenter of $APQ$ and the result follows. For the second part there's no need to bash ,coordinate, sine law or similar stuffs ! Note that $A$ is the orthocenter of $PIQ$ . It's well known [for ex. INMO 2015 ,1 generalised !!] that circumcenter of $PIQ$ lies on $BC$ , it's the in touch point on $BC$ and if $R$ is the circumcenter then $PRQ$ is isosceles right angles triangle ! Then if $X$ is the feet of perpendicular from $R$ to $PQ$ then $AI = 2.PR = PQ$ . The first eq. for orthocenter, circumcenter reasones while the second for isosceles right angles triangle reason !!

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Next observe that $$\angle AQB = \angle APC = 90^{\circ} + \frac{90^{\circ}}{2}=135^{\circ} \implies APG = \angle AQH = 45^{\circ}.$$Now observe that $$\angle HAG = \angle IAQ+\angle IAP = \frac{45^{\circ}}{2}+\frac{45^{\circ}}{2} = 45^{\circ} \implies QI \perp AP\text{ and } PI\perp AQ \implies I\text{ is the orthocentre of }\triangle APQ \implies AI \perp PQ.$$Now recall that if $H$ is the orthocenter of $\triangle ABC$, then $AH=2R\cos A$, thus here $AI = 2R\cos (\angle PAQ)=2R\cos 45^{\circ}$ where $R$ is the radius of $\odot(APQ)$, now note that $\cos 45^{\circ}=\sin 45^{\circ}$ also by extended sine-rule in $\triangle PAQ$ we have $\frac{PQ}{\sin (\angle PAQ)}=\frac{PQ}{\sin 45^{\circ}}=2R$ combining these two get $AI = PQ.$$\quad \blacksquare$

First part: Clearly $B-P-I$ and $C-Q-I$ are collinear. Let $X = QI \cap AP$ and $Y = PI \cap AQ$ $\angle BAI = 45^\circ$, let $\angle IAY = x \implies \angle YAC = 45^\circ - x$, $\angle BAX = \angle XAD = x \implies \angle QCA = \angle QCB = x$ $\implies \angle CBY = \angle YBA = 45^\circ - x$ $\implies \angle YBA + \angle BAY = 45^\circ - x + 45^\circ + x = 90^\circ$ $\implies PY \perp AQ$ similarly $QX \perp AP$ So $I$ is the orthocenter of $\triangle PAQ \implies AI \perp PQ$ Second part: We have triangle $APQ$ with $\angle PAY = 45^\circ = \angle YPA = \angle XIP = \angle YIQ = \angle IQY$ Setting $AX = a$ and $AY = b$ we get that $YQ = a\sqrt 2 - b = YI \implies AI^2 = b^2 + (a\sqrt 2 - b)^2 = 2a^2 + 2b^2 - 2\sqrt 2 ab$ and $PQ^2 = AP^2 + AQ^2 - 2 AP \cdot AQ \cdot \cos \angle PAQ = (b\sqrt 2)^2 + (a \sqrt 2)^2 - 4ab \cos 45^\circ = 2a^2 + 2b^2 - 2\sqrt 2 ab = AI^2$ and we are done.

Part1 : $AI = PQ$. Claim: $DPQ$ and $BAC$ are similar. Proof : Note that $ADB$ and $CDA$ are similar so $\frac{PD}{AB} = \frac{QD}{AC}$ and also $\angle PDQ = \angle 90 = \angle BAC$. Now we have $\frac{PQ}{BC} = \frac{DP}{AB} = \frac{AI}{BC}$ so $PQ = AI$. Part2 : $AI \perp PQ$. Let $QI$ meet $AP$ at $S$ and $PI$ meet $AQ$ at $K$. Claim: $AKDB$ and $ASDC$ are cyclic. Proof : $\angle DBK = \angle DBA/2 = \angle DAC/2 = \angle DAK$ so $AKDB$ is cyclic. we prove the other one with same approach. this implies that $I$ is orthocenter of $APQ$ so $AI \perp PQ$. we're Done.

Clean coordinate bash. Toss the figure on the coordinate plane. Set $A \equiv (0,0),C \equiv (1,0)$ and $B \equiv (0,a)$. It is well known that $D \equiv \left(\dfrac{a^{2}}{a^{2}+1},\dfrac{a}{a^{2}+1}\right)$. Now using the formula of incenter coordinates in triangles $ABD$,$ACD$ and $ABC$ we get: $$ I \equiv \left(\dfrac{a}{a+1+\sqrt{a^{2}+1}},\dfrac{a}{a+1+\sqrt{a^{2}+1}}\right).$$ $$ P \equiv \left(\dfrac{a^{2}}{\sqrt{a^{2}+1}(\sqrt{a^{2}+1}+a+1)},\dfrac{a(1+\sqrt{a^{2}+1})}{\sqrt{a^{2}+1}(\sqrt{a^{2}+1}+a+1)}\right).$$ $$ Q \equiv \left(\dfrac{a(a+\sqrt{a^{2}+1})}{\sqrt{a^{2}+1}(\sqrt{a^{2}+1}+a+1)},\dfrac{a}{\sqrt{a^{2}+1}(\sqrt{a^{2}+1}+a+1)}\right).$$ Clearly $m_{AI}=1$ and $m_{PQ}=\dfrac{a\sqrt{a^{2}+1}}{-a\sqrt{a^{2}+1}}=-1$, hence $AI$ is perpendicular to $PQ$. Now by distance formula one can readily see that, $$AI=PQ=\left(\dfrac{a\sqrt{2}}{a+1+\sqrt{a^{2}+1}}\right).$$ Hence $AI=PQ$ as desired. (QED) $\blacksquare$.