题目（七年级 · 度量几何学）：如图, $B C / / O A, \angle B=\angle A=100^{\circ}$, 点 $E 、 F$ 在 $B C$ 上, $O E$ 平分 $\angle B O F$, 且 $O C$ 平分 $\angle A O F$,下列结论中正确的是 $\qquad$


(1) $\angle F O C=\angle F C O$;
(2) $O B / / A C$;
(3) $\angle E O C=45^{\circ}$;
(4) $\angle O C B: \angle O F B=1: 3$;
(5) 若 $\angle O E B=\angle O C A$, 则 $\angle O C A=60^{\circ}$.
选项：
答案：【答案】(1)(2)

解: (1) $\because B C / / O A, O C$ 平分 $\angle A O F$,

$\therefore \angle F O C=\angle A O C, \angle A O C=\angle F C O$,

则 $\angle F O C=\angle F C O$, 故(1)正确;

(2) $\because B C / / O A$,

$\therefore \angle B+\angle A O B=180^{\circ}$, (两直线平行, 同旁内角互补)

$\because \angle A=\angle B$,

$\therefore \angle A+\angle A O B=180^{\circ}$, (等量代换)

$\therefore O B / / A C$. (同旁内角互补, 两直线平行)

故(2)正确;

(3) $\because \angle A=\angle B=100^{\circ}$,

由(2) 得 $\angle B O A=180^{\circ}-\angle B=80^{\circ}$;

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图(1)

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图(2)

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图(3)

$\because \angle F O C=\angle A O C$, 并且 $O E$ 平分 $\angle B O F$,

$\therefore \angle E O F=\frac{1}{2} \angle B O F, \angle F O C=\frac{1}{2} \angle F O A$,

$\therefore \angle E O C=\angle E O F+\angle F O C=\frac{1}{2}(\angle B O F+\angle F O A)=\frac{1}{2} \angle B O A=40^{\circ}$.

故(3)错误;

(4) $\because B C / / O A$,
$\therefore \angle F C O=\angle C O A$,

又 $\because \angle F O C=\angle A O C$,

$\therefore \angle F O C=\angle F C O$,

$\therefore \angle O F B=\angle F O C+\angle F C O=2 \angle O C B$,

$\therefore \angle O C B: \angle O F B=1: 2$.

故(4)错误;

(5) $\because O B / / A C$,

$\therefore \angle O C A=\angle B O C$,

设 $\angle B O E=\angle E O F=\alpha, \angle F O C=\angle C O A=\beta$,

$\therefore \angle O C A=\angle B O C=2 \alpha+\beta$,

$\angle O E B=\angle E O C+\angle E C O=\alpha+\beta+\beta=\alpha+2 \beta$,

$\because \angle O E B=\angle O C A$,

$\therefore 2 \alpha+\beta=\alpha+2 \beta$,

$\therefore \alpha=\beta$,

$\because \angle A O B=80^{\circ}$,

$\therefore \alpha=\beta=20^{\circ}$,

$\therefore \angle O C A=2 \alpha+\beta=40^{\circ}+20^{\circ}=60^{\circ}$.

故(5)正确.

故答案为(1)(2).
解析：【答案】(1)(2)

解: (1) $\because B C / / O A, O C$ 平分 $\angle A O F$,

$\therefore \angle F O C=\angle A O C, \angle A O C=\angle F C O$,

则 $\angle F O C=\angle F C O$, 故(1)正确;

(2) $\because B C / / O A$,

$\therefore \angle B+\angle A O B=180^{\circ}$, (两直线平行, 同旁内角互补)

$\because \angle A=\angle B$,

$\therefore \angle A+\angle A O B=180^{\circ}$, (等量代换)

$\therefore O B / / A C$. (同旁内角互补, 两直线平行)

故(2)正确;

(3) $\because \angle A=\angle B=100^{\circ}$,

由(2) 得 $\angle B O A=180^{\circ}-\angle B=80^{\circ}$;



图(1)



图(2)



图(3)

$\because \angle F O C=\angle A O C$, 并且 $O E$ 平分 $\angle B O F$,

$\therefore \angle E O F=\frac{1}{2} \angle B O F, \angle F O C=\frac{1}{2} \angle F O A$,

$\therefore \angle E O C=\angle E O F+\angle F O C=\frac{1}{2}(\angle B O F+\angle F O A)=\frac{1}{2} \angle B O A=40^{\circ}$.

故(3)错误;

(4) $\because B C / / O A$,
$\therefore \angle F C O=\angle C O A$,

又 $\because \angle F O C=\angle A O C$,

$\therefore \angle F O C=\angle F C O$,

$\therefore \angle O F B=\angle F O C+\angle F C O=2 \angle O C B$,

$\therefore \angle O C B: \angle O F B=1: 2$.

故(4)错误;

(5) $\because O B / / A C$,

$\therefore \angle O C A=\angle B O C$,

设 $\angle B O E=\angle E O F=\alpha, \angle F O C=\angle C O A=\beta$,

$\therefore \angle O C A=\angle B O C=2 \alpha+\beta$,

$\angle O E B=\angle E O C+\angle E C O=\alpha+\beta+\beta=\alpha+2 \beta$,

$\because \angle O E B=\angle O C A$,

$\therefore 2 \alpha+\beta=\alpha+2 \beta$,

$\therefore \alpha=\beta$,

$\because \angle A O B=80^{\circ}$,

$\therefore \alpha=\beta=20^{\circ}$,

$\therefore \angle O C A=2 \alpha+\beta=40^{\circ}+20^{\circ}=60^{\circ}$.

故(5)正确.

故答案为(1)(2).