<Tri-2025-2295>, hint

Suppose A altitude(AD), B-bisector (BP)and C-median(CM) meet at  H. 

By Ceva's (AM/MB ) x(BD/DC)x(CP/PA)=1 

(1/1)x(C cosB)/BcosC) x(a/c)=1 (sinCcosB)/(sinBcosC)=sinC/sinA

sinA cosB=sinB cosC...(1)   If A=B=C=60 (1) is right

Suppose A=90, then cosB=sinB sinB  cos^2B+cosB-1=0,  Hence cosB=(r5-1)/2