试题分析:根据点F是底角平分线的交点,可得点F是三角形ABC角平分线的交点,连接AF,则AF平分∠BAC,设∠C=x,利用等腰三角形的性质分别得出∠BAF、∠ABF、∠AFB,然后利用三角形的内角和定理可得出答案.
如图,连接AF,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230225460855439.png)
∵点F是底角平分线的交点,
∴点F是三角形ABC角平分线的交点(三角形的额角平分线交于一点),
∴AF平分∠BAC,
设∠C=x,则∠ABF=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823022546085339.png)
x,∠BAF=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823022546085339.png)
∠BAC=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823022546085339.png)
(180°-2x)=90°-x,
又∵BF=DF,AD=DF(折叠的性质),
∴∠FDB=∠FBD,∠DAF=∠DFA,
∴∠DFB=180°-2∠ABF=180°-x,
∴∠AFB=∠DFB+∠AFD=∠DFB+∠DAF=180°-x+(90°-x)=270°-2x,
在三角形ABF中,∠BAF+∠ABF+∠AFB=180°,即(90°-x)+(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823022546085339.png)
x)+(270°-2x)=180°,
解得:x=72°,即∠C=72°.
故选C.
点评:解答本题的关键是熟练掌握三角形的三条角平分线相交于一点.