4. 如图5,在$\mathrm{Rt}△ ABC$中,$∠ BAC=90°$,分别以点$C$,$B$为圆心,以大于$\dfrac{1}{2}BC$为半径画弧,两弧相交于点$M$,$N$,作直线$MN$分别交$AB$,$CB$于点$D$,$E$,连结$CD$,$AE$相交于点$P$. 若$∠ B=23°$,则$∠ APC$的大小为.

4. $69°$

三、解答题
1. 如图6,在$\mathrm{Rt}△ ABC$中,$∠ ACB=90°$,$∠ A=30°$,点$D$是$AB$的中点,点$E$为边$AC$上一点,连结$CD$,$DE$,以$DE$为边在$DE$的左侧作等边三角形$DEF$,连结$BF$.
(1)求证:$△ BCD$为等边三角形;

(2)求证:$∠ DBF=∠ DCE$.

1. (1)证明:在$\mathrm{Rt}△ ABC$中,$∠ ACB=90°$,$∠ A=30°$,$\therefore∠ ABC=60°$,$\because$点$D$是$AB$的中点,$\therefore CD=BD=AD=\dfrac{1}{2}AB$,$\therefore△ BCD$为等边三角形 (2)提示:证明$△ BDF≌△ CDE$即可

2. 如图7-1,在矩形$ABCD$中,过矩形$ABCD$对角线$AC$的中点$O$作$EF⊥ AC$分别交$AB$、$DC$于$E$、$F$点.

(1)求证:$AE = CF$;
(2)如图7-2,若$G$为$AE$的中点,且$∠ AOG=30°$,求证:$DC=3OG$.

2. (1)提示:证明$△ COF≌△ AOE$即可 (2)证明:连结$OB$,$\because ABCD$是矩形,$O$是$AC$中点,$\therefore OA=OB$,在$\mathrm{Rt}△ AOE$中,$G$为$AE$中点,$\therefore OG=AG=GE$,又$\because∠ AOG=30°$,$\therefore∠ OAG=30°$,$∠ OEA=60°$,$\therefore OG=OE=GE$,$\therefore∠ OBA=∠ OAG=30°$,$\therefore∠ EOB=∠ OEA-∠ OBA=30°=∠ OBE$,$\therefore OE=EB$,$\therefore OG=AG=GE=BE$,$\therefore AB=3OG$,即$DC=3OG$