23. (12分)【初步认识】
(1)如图1,线段AB,CD相交于点O,连接AD,BC.求证:∠A + ∠D = ∠B + ∠C;
【继续探索】
(2)如图2,∠A = m°,∠C = n°,∠ABC,∠ADC的平分线BP,DP相交于点P.
①用m,n表示∠P的度数为;
②若m = 40,n = 32,求∠P的度数;
(3)如图3,∠ABC,∠ADC的平分线BP,DP相交于点P,∠DAB,∠DCB的平分线AQ,CQ相交于点Q.若∠P = ∠Q,判断AD与BC的位置关系并说明理由.
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本题考查三角形的内角和定理,角平分线的性质,平行线的判定.
解析(1)证明:
∵在△AOD中,∠A + ∠D + ∠AOD = 180°,
∴∠A + ∠D = 180° - ∠AOD.
又
∵在△BOC中,∠B + ∠C + ∠BOC = 180°,
∴∠B + ∠C = 180° - ∠BOC.
又
∵∠AOD = ∠BOC,
∴∠A + ∠D = ∠B + ∠C.
(2)①同(1)可得,∠A + ∠ADC = ∠ABC + ∠C,∠A + ∠ADP = ∠P + ∠ABP.
∵BP平分∠ABC,DP平分∠ADC,
∴∠ABP = $\frac{1}{2}$∠ABC,∠ADP = $\frac{1}{2}$∠ADC,
∴∠A + $\frac{1}{2}$∠ADC = ∠P + $\frac{1}{2}$∠ABC,
∴2∠A + ∠ADC = 2∠P + ∠ABC.
又
∵∠A + ∠ADC = ∠ABC + ∠C,
∴∠A = 2∠P - ∠C,
∴∠P = $\frac{∠A + ∠C}{2}$.
∵∠A = m°,∠C = n°,
∴∠P = $(\frac{m + n}{2})°$. 故答案为$(\frac{m + n}{2})°$.
②
∵m = 40,n = 32,
∴∠P = $(\frac{m + n}{2})° = \frac{40° + 32°}{2} = 36°$.
(3)AD//BC. 理由如下:
同(2)可得∠P = $\frac{∠DAB + ∠DCB}{2}$,∠Q = $\frac{∠ABC + ∠ADC}{2}$.
∵∠P = ∠Q,
∴$\frac{∠DAB + ∠DCB}{2} = \frac{∠ABC + ∠ADC}{2}$,
∴∠DAB + ∠DCB = ∠ABC + ∠ADC.
又
∵∠DAB + ∠ADC = ∠DCB + ∠ABC,
∴2∠DAB + ∠DCB + ∠ADC = 2∠ABC + ∠DCB + ∠ADC,
∴∠DAB = ∠ABC,
∴AD//BC.