4. 如图，已知$FC// AB// DE$，H为FC上一点，若$∠ BHD:∠ D:∠ B=2:3:4$，则$∠ D$的度数为
108°
.答案：4. $108°$
5. 如图，直线AB与CD相交于点O，$OE⊥ AB$.
(1)若$∠ EOC:∠ COA=7:3$，求$∠ BOD$的度数.
(2)从点O出发在$∠ AOD$内部引射线OF，若$∠ AOC$与$∠ EOF$互补，则OD与OF有什么位置关系？并说明理由.
(1)若$∠ EOC:∠ COA=7:3$，求$∠ BOD$的度数.
(2)从点O出发在$∠ AOD$内部引射线OF，若$∠ AOC$与$∠ EOF$互补，则OD与OF有什么位置关系？并说明理由.
答案：5.解:(1)$\because OE⊥ AB$,$\therefore ∠ AOE=90°$.
$\because ∠ EOC:∠ COA=7:3$,
$\therefore$设$∠ COA=3x$,则$∠ EOC=7x$,
$\therefore 7x+3x=90°$,$\therefore x=9°$,$\therefore ∠ AOC=27°$,
$\therefore ∠ BOD=∠ AOC=27°$.
(2)$OD⊥ OF$.理由如下:
$\because ∠ AOC$与$∠ EOF$互补,
$\therefore ∠ AOC+∠ EOF=180°$.
$\because ∠ AOC+∠ AOD=180°$,$\therefore ∠ AOD=∠ EOF$,
$\therefore ∠ AOD-∠ AOF=∠ EOF-∠ AOF$,
$\therefore ∠ DOF=∠ AOE=90°$,$\therefore OF⊥ OD$.
$\because ∠ EOC:∠ COA=7:3$,
$\therefore$设$∠ COA=3x$,则$∠ EOC=7x$,
$\therefore 7x+3x=90°$,$\therefore x=9°$,$\therefore ∠ AOC=27°$,
$\therefore ∠ BOD=∠ AOC=27°$.
(2)$OD⊥ OF$.理由如下:
$\because ∠ AOC$与$∠ EOF$互补,
$\therefore ∠ AOC+∠ EOF=180°$.
$\because ∠ AOC+∠ AOD=180°$,$\therefore ∠ AOD=∠ EOF$,
$\therefore ∠ AOD-∠ AOF=∠ EOF-∠ AOF$,
$\therefore ∠ DOF=∠ AOE=90°$,$\therefore OF⊥ OD$.
6. 如图，$AB// CD$，$∠ BCD$的平分线CE交BD于点E，连接AE.若$∠ DBC=2∠ ABC$，$∠ BDC=6∠ BAE$，求$∠ AEC$的度数.
答案：
6.解:过点$E$作$EF// AB$,如答图.
$\because AB// CD$,
$\therefore AB// CD// EF$,
$\therefore ∠ BAE=∠ AEF$,$∠ DCE=∠ CEF$,$∠ ABC=∠ BCD$,
$\therefore ∠ AEC=∠ AEF+∠ CEF=∠ BAE+∠ DCE$.
$\because ∠ BCD$的平分线$CE$交$BD$于点$E$,
$\therefore ∠ DCE=∠ BCE$,
设$∠ DCE=∠ BCE=x$,则$∠ ABC=∠ BCD=2x$,
$\therefore ∠ DBC=2∠ ABC=4x$.
设$∠ BAE=y$,则$∠ BDC=6∠ BAE=6y$.
$\because AB// CD$,$\therefore ∠ ABD+∠ BDC=180°$,
$\therefore 2x+4x+6y=180°$,解得$x+y=30°$,
$\therefore ∠ AEC=∠ BAE+∠ DCE=y+x=30°$.
6.解:过点$E$作$EF// AB$,如答图.
$\because AB// CD$,
$\therefore AB// CD// EF$,
$\therefore ∠ BAE=∠ AEF$,$∠ DCE=∠ CEF$,$∠ ABC=∠ BCD$,
$\therefore ∠ AEC=∠ AEF+∠ CEF=∠ BAE+∠ DCE$.
$\because ∠ BCD$的平分线$CE$交$BD$于点$E$,
$\therefore ∠ DCE=∠ BCE$,
设$∠ DCE=∠ BCE=x$,则$∠ ABC=∠ BCD=2x$,
$\therefore ∠ DBC=2∠ ABC=4x$.
设$∠ BAE=y$,则$∠ BDC=6∠ BAE=6y$.
$\because AB// CD$,$\therefore ∠ ABD+∠ BDC=180°$,
$\therefore 2x+4x+6y=180°$,解得$x+y=30°$,
$\therefore ∠ AEC=∠ BAE+∠ DCE=y+x=30°$.
7. 如图，$AB⊥ BC$，$DC⊥ BC$，E是BC上一点，$EM⊥ EN$，$∠ EMA$和$∠ END$的平分线交于点F，求$∠ MFN$的度数.
答案：
7.解:过点$E$作$EH// AB$,过点$F$作$FQ// AB$,如答图,则
$AB// EH// FQ$.
$\because ∠ EMA$和$∠ END$的平分线交于点$F$,
$\therefore ∠ AMF=∠ EMF$,$∠ ENF=∠ FND$.
设$∠ AMF=∠ EMF=x$,$∠ ENF=∠ FND=y$,
$\because AB⊥ BC$,$DC⊥ BC$,$\therefore AB// CD$,
$\therefore AB// EH// FQ// DC$,
$\therefore ∠ MEH=∠ BME=180°-2x$,$∠ CNE=∠ HEN=$
$180°-2y$,$∠ QFM=∠ AMF=x$,$∠ QFN=∠ FND=y$.
$\because EM⊥ EN$,$\therefore ∠ MEN=90°$.
又$\because ∠ MEN=∠ MEH+∠ HEN$,
$\therefore 180°-2x+180°-2y=90°$,解得$x+y=135°$,
$\therefore ∠ MFN=∠ MFQ+∠ NFQ=x+y=135°$.
7.解:过点$E$作$EH// AB$,过点$F$作$FQ// AB$,如答图,则
$AB// EH// FQ$.
$\because ∠ EMA$和$∠ END$的平分线交于点$F$,
$\therefore ∠ AMF=∠ EMF$,$∠ ENF=∠ FND$.
设$∠ AMF=∠ EMF=x$,$∠ ENF=∠ FND=y$,
$\because AB⊥ BC$,$DC⊥ BC$,$\therefore AB// CD$,
$\therefore AB// EH// FQ// DC$,
$\therefore ∠ MEH=∠ BME=180°-2x$,$∠ CNE=∠ HEN=$
$180°-2y$,$∠ QFM=∠ AMF=x$,$∠ QFN=∠ FND=y$.
$\because EM⊥ EN$,$\therefore ∠ MEN=90°$.
又$\because ∠ MEN=∠ MEH+∠ HEN$,
$\therefore 180°-2x+180°-2y=90°$,解得$x+y=135°$,
$\therefore ∠ MFN=∠ MFQ+∠ NFQ=x+y=135°$.