Let $M_a, M_b, M_c$ are midpoints of segmets $BC, CA, AB$ respectively, $AP, BQ, CR$ are altitudes from $A, B, C$ respectively, $\Omega$ is circle with diameter $BC$. It's obviously thet $R, Q \in \Omega$. If bisectors of $\angle ABH$ and $\angle ACH$ intersect $\Omega$ in points $A_{1-1}$ and $A_{1-2}$, then $A_{1-1}R=A_{1-1}Q$ and $A_{1-2}R=A_{1-2}Q$ $\implies$ $A_{1-1}=A_{1-2}=A_1$. Then points $A_1, E, M_a$ lie on a perpendicular bisector to line $QR$, and $EA_1=M_aA_1-EM_a=\frac{1}{2}BC-\frac{1}{2}R$ $\implies$ $EA_1+EB_1+EC_1=p-\frac{3R}{2}$. We are done!