10. (2024·泗洪县期末)如图，在△ABC中，AD是△ABC的角平分线，P为线段AD上的一个动点，PE⊥AD交直线BC于点E.
(1)若∠B=30°，∠ACB=80°，求∠E的度数；
(2)当点P在线段AD上运动时，若∠E是锐角，请直接写出∠E，∠ACB，∠B之间的关系并说明理由.
答案：10. 解:(1)因为∠B = $30^{\circ}$,∠ACB = $80^{\circ}$,所以∠BAC = $70^{\circ}$.因为AD平分∠BAC,所以∠DAC = $35^{\circ}$,所以∠ADC = $65^{\circ}$,所以∠E = $25^{\circ}$.
(2)∠E = $\frac{1}{2}$(∠ACB - ∠B).理由如下:
如答图,设∠B = $n^{\circ}$,∠ACB = $m^{\circ}$.
因为AD平分∠BAC,所以∠1 = ∠2 = $\frac{1}{2}$∠BAC;
因为∠B + ∠ACB + ∠BAC = $180^{\circ}$,∠B = $n^{\circ}$,∠ACB = $m^{\circ}$,所以∠CAB = $(180 - n - m)^{\circ}$,所以∠BAD = $\frac{1}{2}$(180 - n - m)°,
所以∠3 = ∠B + ∠1 = $n^{\circ} + \frac{1}{2}(180 - n - m)^{\circ} = 90^{\circ} + \frac{1}{2}n^{\circ} - \frac{1}{2}m^{\circ}$.
因为PE⊥AD,所以∠DPE = $90^{\circ}$,
所以∠E = $90^{\circ} - (90^{\circ} + \frac{1}{2}n^{\circ} - \frac{1}{2}m^{\circ}) = \frac{1}{2}(m^{\circ} - n^{\circ}) = \frac{1}{2}$(∠ACB - ∠B).
