6. (2024·南京期末)如图,直线$AB,CD$相交于点$O$,$OE$平分$∠ BOD$.
(1)若$∠ AOC=72°$,求$∠ COE$的度数;
(2)若$OF$平分$∠ AOE$,$∠ DOF=54°$,求$∠ AOC$的度数.
答案：6.解:(1)$\because ∠AOC=72^{\circ },\therefore ∠BOD=72^{\circ }.$
$\because OE$平分$∠BOD,\therefore ∠EOD=\frac {1}{2}∠BOD=36^{\circ },$
$\therefore ∠COE=180^{\circ }-36^{\circ }=144^{\circ }.$
(2)如答图,$\because OE$平分$∠BOD,\therefore ∠1=∠2.$
$\because OF$平分$∠AOE,\therefore ∠AOF=∠EOF.$
$\because ∠DOF=54^{\circ },$
$\therefore ∠AOF=∠EOF=∠2+∠DOF=∠2+54^{\circ }.$
$\because ∠AOF+∠EOF+∠1=180^{\circ },$
$\therefore ∠2+54^{\circ }+∠2+54^{\circ }+∠2=180^{\circ },$
$\therefore ∠2=24^{\circ },\therefore ∠1=∠2=24^{\circ },$
$\therefore ∠AOC=∠BOD=∠1+∠2=48^{\circ }.$
