In $\triangle ABC$, the perpendiculars $AD, BE, CF$ from the vertices to the opposite sides intersect at $O$. Then —

• a$AO + OD = BO + OE = CO + OF$
• b$AO \cdot OD = BO \cdot OE = CO \cdot OF$
• c$\dfrac{AO}{OD} = \dfrac{BO}{OE} = \dfrac{CO}{OF}$
• d$AO = OD$, $BO = OE$, $CO = OF$