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【典例1】在$△ ABC$中，$BD$平分$∠ ABC$，$CE$平分$∠ ACB$，$BD$与$CE$交于点$M$。
(1)如图，若$MN⊥ BC$于$N$，$∠ A=60°$，求$∠1-∠2$的值；
(2)若$∠ BEC=x$，$∠ BDC=y$，那么$∠ BMC=$
(1)如图，若$MN⊥ BC$于$N$，$∠ A=60°$，求$∠1-∠2$的值；
(2)若$∠ BEC=x$，$∠ BDC=y$，那么$∠ BMC=$
$60°+\dfrac{x+y}{3}$
。答案
(1) 解:设$∠ ABD=α,∠ ACE=β$,则$∠ 2=α,∠ 1=90°-β$,
$\because ∠ A+2α+2β=180°,\therefore α+β=60°$,
$\therefore ∠ 1-∠ 2=30°$;
(2) $\because x+2α+β=180°,y+α+2β=180°$,
$\therefore x+y+3(α+β)=360°$,
而$∠ BMC=180°-(α+β)$,
$\therefore ∠ BMC=60°+\dfrac{x+y}{3}$.
$\because ∠ A+2α+2β=180°,\therefore α+β=60°$,
$\therefore ∠ 1-∠ 2=30°$;
(2) $\because x+2α+β=180°,y+α+2β=180°$,
$\therefore x+y+3(α+β)=360°$,
而$∠ BMC=180°-(α+β)$,
$\therefore ∠ BMC=60°+\dfrac{x+y}{3}$.
变式.(2026·武昌)如图,已知点P为△ABC三条内角平分线AD,BE,CF的交点,作DG⊥PC于G,则∠PDG等于(
A.∠ABE
B.∠DAC
C.∠BCF
D.∠CPE
A
)A.∠ABE
B.∠DAC
C.∠BCF
D.∠CPE
答案
解:设$∠ ACF=α,∠ PAC=x,∠ ABE=β$,
$\because 2α+2x+2β=180°,\therefore α+x+β=90°$,
又$\because ∠ DPG=α+x=90°-β$,
$\therefore ∠ PDG=90°-∠ DPG=β=∠ ABP$.
$\because 2α+2x+2β=180°,\therefore α+x+β=90°$,
又$\because ∠ DPG=α+x=90°-β$,
$\therefore ∠ PDG=90°-∠ DPG=β=∠ ABP$.
【典例2】如图，I为$△ ABC$角平分线的交点，过I点作$DE ⊥ AI$，分别交AB，AC于D，E，求证：$∠ BIC=∠ BDI=∠ CEI$.
答案
解:设$∠ IBC=β,∠ ICB=α,∠ IAD=x$,
$\therefore 2α+2β+2x=180°,\therefore α+β+x=90°$,
$\therefore ∠ BIC=180°-(α+β)=90°+x$,
又$\because$在$△ ADI$中,$∠ BDI=90°+x$,
$\therefore ∠ BIC=∠ BDI=∠ CEI$.
$\therefore 2α+2β+2x=180°,\therefore α+β+x=90°$,
$\therefore ∠ BIC=180°-(α+β)=90°+x$,
又$\because$在$△ ADI$中,$∠ BDI=90°+x$,
$\therefore ∠ BIC=∠ BDI=∠ CEI$.
变式.如图，AO，BO分别平分∠CAB，∠CBA，OD⊥OB交AB于点D，求$\frac{∠ AOD}{∠ C}$的值.
答案
解:$∠ AOB=90°+\dfrac{1}{2}∠ C,\therefore \dfrac{∠ AOD}{∠ C}=\dfrac{1}{2}$.
