\bump please!

Here is a sketch of my solution during the contest (back in 2021): Because $(ABE)\perp A'D$ it follows that $BE\perp A'D$, and by using the similar relations we get that $G$ is the orthocentre of $\triangle A'DB$ and similarly that $P$ is the orthocentre of $\triangle BC'D$. $\qquad (1)$ By the reverse of the three perpendiculars theorem it follows that $AG\perp (A'DB)$ and $CP\perp (BC'D)$. $\qquad (2)$ Now, by using $(1)$ we get that $BD\perp (A'AG)$ and that $BD\perp (C'CP)$, which implies $(A'AG)\parallel (C'CP)$. Let $X\in A'G\cap BD, Y\in C'P\cap BD$. By using $a>b$ and some trig we get that $D$, $X$, $Y$ and $B$ are in this order on the line $BD$, which implies \[\mathrm{dist}((A'AG),(C'CP))=XY=BD-2BY=\sqrt{a^2+b^2}-\frac{2b^2}{\sqrt{a^2+b^2}}=\frac{a^2-b^2}{\sqrt{a^2+b^2}}.\]by using the formula of the leg of a right-angled triangle. Now, for the part b, let $(ABC)=\alpha$, $Q=\mathrm{pr}_\alpha (G)$, $R=\mathrm{pr}_\alpha (P)$. From $\triangle A'BD\cong\triangle C'DB$ it follows that $\frac{XG}{XA'}=\frac{YP}{YC'}$ and from $\triangle XGQ\sim\triangle XAA'$ and $\triangle YPR\sim\triangle YC'C$ it follows that $\frac{XG}{XA'}=\frac{GQ}{c}$ and that $\frac{YP}{YC'}=\frac{PR}{c}$, so $GQ=PR$, which implies $GP\parallel \alpha$. Now, we just have to compute $PR$. Here, by using the formulas for the leg and for the height of a right-angled triangle we get \[YC=\frac{ab}{\sqrt{a^2+b^2}},\;YC'=\frac{\sqrt{\sum a^2b^2}}{\sqrt{a^2+b^2}},\; CP=\frac{abc}{\sqrt{\sum a^2b^2}}\text{ and } YP=\frac{a^2b^2}{\sqrt{a^2+b^2}\sqrt{\sum a^2b^2}},\]which implies \[PR=\frac{PC\cdot PY}{YC}=\frac{a^2b^2c}{a^2b^2+b^2c^2+c^2a^2}.\]Thus, the problem is solved.

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