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14. 观察下列等式：①$2× 4+1=3^{2}$；②$4× 6+1=5^{2}$；③$6× 8+1=7^{2}$；$···$。
(1)根据你发现的规律，请写出第$4$个等式：
(2)根据你发现的规律，请写出第$n$($n$为正整数)个等式，并证明你所写出的等式的正确性。
(3)请写出第$198$个等式：
(1)根据你发现的规律，请写出第$4$个等式：
$8×10 + 1 = 9^{2}$
。(2)根据你发现的规律，请写出第$n$($n$为正整数)个等式，并证明你所写出的等式的正确性。
(3)请写出第$198$个等式：
$396×398 + 1 = 397^{2}$
。答案
14. (1)$8×10 + 1 = 9^{2}$ (2)第n(n为正整数)个等式为$2n(2n + 2)+1=(2n + 1)^{2}$。证明如下:左边$=2n(2n + 2)+1 = 4n^{2}+4n + 1$，右边$=(2n + 1)^{2}=4n^{2}+4n + 1$，$\therefore$左边 = 右边，$\therefore$等式成立。 (3)$396×398 + 1 = 397^{2}$
15. 在四边形$ABCD$中，$∠ A=90^{\circ}$，$∠ ABC$、$∠ ADC$的平分线分别交直线$CD$、$AB$于点$E$、$F$。
(1)如图$1$，若$∠ C=90^{\circ}$，求证：$EB// DF$。
(2)如图$2$，若线段$DF$、$EB$交于点$P$，$∠ BPF=20^{\circ}$，求$∠ C$的度数。
(1)如图$1$，若$∠ C=90^{\circ}$，求证：$EB// DF$。
(2)如图$2$，若线段$DF$、$EB$交于点$P$，$∠ BPF=20^{\circ}$，求$∠ C$的度数。
答案
15. (1)证明:$\because ∠A = ∠C = 90^{\circ }$，$∠A + ∠ABC + ∠C + ∠ADC = 360^{\circ }$，$\therefore ∠ABC + ∠ADC = 360^{\circ } - 90^{\circ } - 90^{\circ } = 180^{\circ }$。$\because BE$平分$∠ABC$，$DF$平分$∠ADC$，$\therefore ∠ABC = 2∠ABE$，$∠ADC = 2∠ADF$，$\therefore 2∠ABE + 2∠ADF = 180^{\circ }$，$\therefore ∠ABE + ∠ADF = 90^{\circ }$。$\because ∠A = 90^{\circ }$，$\therefore ∠AFD + ∠ADF = 90^{\circ }$，$\therefore ∠AFD = ∠ABE$，$\therefore EB// DF$。 (2)$\because BE$平分$∠ABC$，$DF$平分$∠ADC$，$\therefore ∠ABE=\frac {1}{2}∠ABC$，$∠ADF=\frac {1}{2}∠ADC$。$\because ∠BFP$是$△ ADF$的外角，$∠A = 90^{\circ }$，$\therefore ∠BFP = ∠A + ∠ADF = 90^{\circ }+\frac {1}{2}∠ADC$。$\because ∠BFP + ∠BPF + ∠ABE = 180^{\circ }$，$∠BPF = 20^{\circ }$，$\therefore 90^{\circ }+\frac {1}{2}∠ADC + 20^{\circ }+\frac {1}{2}∠ABC = 180^{\circ }$，$\therefore ∠ADC + ∠ABC = 140^{\circ }$。$\because ∠A + ∠ABC + ∠C + ∠ADC = 360^{\circ }$，$\therefore 90^{\circ } + 140^{\circ } + ∠C = 360^{\circ }$，$\therefore ∠C = 130^{\circ }$。
解析
(1)证明：$\because ∠ A = ∠ C = 90^{\circ}$，四边形内角和为$360^{\circ}$，
$\therefore ∠ ABC + ∠ ADC = 360^{\circ} - 90^{\circ} - 90^{\circ} = 180^{\circ}$。
$\because BE$平分$∠ ABC$，$DF$平分$∠ ADC$，
$\therefore ∠ ABC = 2∠ ABE$，$∠ ADC = 2∠ ADF$，
$\therefore 2∠ ABE + 2∠ ADF = 180^{\circ}$，即$∠ ABE + ∠ ADF = 90^{\circ}$。
$\because ∠ A = 90^{\circ}$，在$△ ADF$中，$∠ AFD + ∠ ADF = 90^{\circ}$，
$\therefore ∠ AFD = ∠ ABE$，
$\therefore EB // DF$。
(2)$\because BE$平分$∠ ABC$，$DF$平分$∠ ADC$，
$\therefore ∠ ABE = \frac{1}{2}∠ ABC$，$∠ ADF = \frac{1}{2}∠ ADC$。
$\because ∠ BFP$是$△ ADF$的外角，$∠ A = 90^{\circ}$，
$\therefore ∠ BFP = ∠ A + ∠ ADF = 90^{\circ} + \frac{1}{2}∠ ADC$。
在$△ BFP$中，$∠ BFP + ∠ BPF + ∠ ABE = 180^{\circ}$，且$∠ BPF = 20^{\circ}$，
$\therefore 90^{\circ} + \frac{1}{2}∠ ADC + 20^{\circ} + \frac{1}{2}∠ ABC = 180^{\circ}$，
$\therefore \frac{1}{2}(∠ ADC + ∠ ABC) = 70^{\circ}$，即$∠ ADC + ∠ ABC = 140^{\circ}$。
$\because$四边形内角和为$360^{\circ}$，$∠ A = 90^{\circ}$，
$\therefore ∠ C = 360^{\circ} - ∠ A - (∠ ABC + ∠ ADC) = 360^{\circ} - 90^{\circ} - 140^{\circ} = 130^{\circ}$。
$\therefore ∠ ABC + ∠ ADC = 360^{\circ} - 90^{\circ} - 90^{\circ} = 180^{\circ}$。
$\because BE$平分$∠ ABC$，$DF$平分$∠ ADC$，
$\therefore ∠ ABC = 2∠ ABE$，$∠ ADC = 2∠ ADF$，
$\therefore 2∠ ABE + 2∠ ADF = 180^{\circ}$，即$∠ ABE + ∠ ADF = 90^{\circ}$。
$\because ∠ A = 90^{\circ}$，在$△ ADF$中，$∠ AFD + ∠ ADF = 90^{\circ}$，
$\therefore ∠ AFD = ∠ ABE$，
$\therefore EB // DF$。
(2)$\because BE$平分$∠ ABC$，$DF$平分$∠ ADC$，
$\therefore ∠ ABE = \frac{1}{2}∠ ABC$，$∠ ADF = \frac{1}{2}∠ ADC$。
$\because ∠ BFP$是$△ ADF$的外角，$∠ A = 90^{\circ}$，
$\therefore ∠ BFP = ∠ A + ∠ ADF = 90^{\circ} + \frac{1}{2}∠ ADC$。
在$△ BFP$中，$∠ BFP + ∠ BPF + ∠ ABE = 180^{\circ}$，且$∠ BPF = 20^{\circ}$，
$\therefore 90^{\circ} + \frac{1}{2}∠ ADC + 20^{\circ} + \frac{1}{2}∠ ABC = 180^{\circ}$，
$\therefore \frac{1}{2}(∠ ADC + ∠ ABC) = 70^{\circ}$，即$∠ ADC + ∠ ABC = 140^{\circ}$。
$\because$四边形内角和为$360^{\circ}$，$∠ A = 90^{\circ}$，
$\therefore ∠ C = 360^{\circ} - ∠ A - (∠ ABC + ∠ ADC) = 360^{\circ} - 90^{\circ} - 140^{\circ} = 130^{\circ}$。
