solution: make high $CD \perp AB$,since it is acute triangle, $D$ will fall on $AB$. $EC \perp CB,$ circle C @$r=CB$, cross $EC$ at $E$, make $EF \perp EC$, extend $CA$, cross $ FE $ at $F$, circle B @$r=BA$, cross line $CB$ at $G$, connect $BF$, make $GH$//$BF$, cross $EF$ at $H$,connect $CH$, make it's midpoint $ M $, circle M@$r=MH$, cross $CG$ at $N$, connect $HN$, make it's midpoint $S$,,circle H @r=HS, cross circle $M$ at $T$, connect $CT$, circle C@$r=CB$, cross $CT$ at $B'$,circle C @$r=CA$, cross $CB$ at $H$, cross $CT$ at $U$,make $\angle B'CA$ bisector $CV$, cross circle $CA$ at $V$, circle V@r=VH, cross circle CA at $A'$, connect $A'B'$, make $RB$//$A'B'$, cross $CT$ at $R$, connect $A'C$,make $AQ$//$A'C$, cross $CT$ at $Q$, cross $BR$ at $P, \triangle PQR $ is the solution. Proof: put $CB$ on $x $, $CE $on y, $C$ is on $O$ $ \implies C(0,0),B(a,0),A(b\cos{C},b\sin{C})$ ,$\angle B'CB=\alpha,B'(a\cos{\alpha},a\sin{\alpha)},A'(b\cos{(\alpha+C)},b\sin{(\alpha+C)})$,$k_{A'B'}=\dfrac{b\sin{(\alpha+C)}-a\sin{\alpha}}{b\cos{(\alpha+C)}-a\sin{\alpha}}$$=\dfrac{b(\sin{C}+\cos{C}\tan{\alpha})-a\tan{\alpha}}{b(\cos{C}-\sin{C}\tan{\alpha})-a},k_{A'C}=\tan{(\alpha+C)}$ line $CB' : y=\tan{\alpha}*x $ line $BR: y=k_{A'B'}(x-a)$ line $AQ: y-b\sin{C}=k_{A'C}(x-b\cos{C})$ let $R(x_1,y_1),Q(x_2,y-2) \implies x_1-x_2=\dfrac{a\cos{\alpha}}{2}$, $x_1=\dfrac{a\cos{\alpha}(b\sin{C}\cos{\alpha}+(b\ \cos{C}-a)\sin{\alpha})}{b\sin{C}} \\ x_2=\dfrac{b\sin{\alpha}\cos{\alpha}}{\sin{C}} \implies \\ (ab\sin{C})\cos{\alpha}+(ab\cos{C}-(a^2+b^2))\sin{\alpha}=\dfrac{ab\sin{C}}{2}$ note: ${S=\dfrac{ab\sin{C}}{2}=\dfrac{ch_c}{2},c^2=a^2+b^2-2ab\cos{C} \implies \cos{\alpha}}-p\sin{\alpha}=\dfrac{1}{2},p=\dfrac{c+\dfrac{h}{\tan{C}}}{h}$ define: $\sin{\theta}=\dfrac{1}{\sqrt{p^2+1}}=\dfrac{h}{(\sqrt{c+\dfrac{h}{\tan{C}})^2+h^2}},\implies \sin{(\theta-\alpha)}=\dfrac{1}{2\sqrt{p^2+1}}$ define : $\sin{D}=\dfrac{1}{2\sqrt{p^2+1}}\implies \alpha=\theta-D$ it is trivial $\triangle ABC = \triangle A'B'C$

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in fact, it is always have the solution for any triangle, because we can always find a side which the high foot fall on the side.that make the construction method works.

bir tinga qimmat masala bu