a) Ta có:
AB=AC (gt)
=> tam giác ABC cân
=>![\angle B= \angle C](http://latex.codecogs.com/gif.latex?%5Cangle%20B%3D%20%5Cangle%20C)
Ta lại có:
![\angle A+\angle B+\angle C = 180^{o}](http://latex.codecogs.com/gif.latex?%5Cangle%20A+%5Cangle%20B+%5Cangle%20C%20%3D%20180%5E%7Bo%7D)
=> ![40^{o}+\angle B+ \angle B =180^{o}](http://latex.codecogs.com/gif.latex?40%5E%7Bo%7D+%5Cangle%20B+%20%5Cangle%20B%20%3D180%5E%7Bo%7D)
=>![2 \angle B =180^{o}-40^{o}=140^{o}](http://latex.codecogs.com/gif.latex?2%20%5Cangle%20B%20%3D180%5E%7Bo%7D-40%5E%7Bo%7D%3D140%5E%7Bo%7D)
=>![\angle B =140^{o}/2=70^{o}](http://latex.codecogs.com/gif.latex?%5Cangle%20B%20%3D140%5E%7Bo%7D/2%3D70%5E%7Bo%7D)
Xét tam giác ABM và tam giác ACM
AB=AC(gt)
MB=MC(M là trung điểm của BC)
Chung cạnh AM
=> tam giác ABM = tam giác ACM
=> ![\angle BAM =\angle CAM](http://latex.codecogs.com/gif.latex?%5Cangle%20BAM%20%3D%5Cangle%20CAM)
![\angle BMA =\angle CMA](http://latex.codecogs.com/gif.latex?%5Cangle%20BMA%20%3D%5Cangle%20CMA)
Ta có:
(kề bù)
=>![\angle BMA+\angle BMA=180^{o}](http://latex.codecogs.com/gif.latex?%5Cangle%20BMA+%5Cangle%20BMA%3D180%5E%7Bo%7D)
=>![2\angle BMA=180^{o}](http://latex.codecogs.com/gif.latex?2%5Cangle%20BMA%3D180%5E%7Bo%7D)
=>
(1)
Ta có :
![\angle BAM +\angle CAM= \angle CAB=40^{o}](http://latex.codecogs.com/gif.latex?%5Cangle%20BAM%20+%5Cangle%20CAM%3D%20%5Cangle%20CAB%3D40%5E%7Bo%7D)
=>![\angle BAM +\angle BAM=40^{o}](http://latex.codecogs.com/gif.latex?%5Cangle%20BAM%20+%5Cangle%20BAM%3D40%5E%7Bo%7D)
=>
=![40^{o}](http://latex.codecogs.com/gif.latex?40%5E%7Bo%7D)
=>
(2)
b) Từ (1)=>AM vuông góc với BC
c) Từ(2)=> AM là tia phân giác ![\angle CAB](http://latex.codecogs.com/gif.latex?%5Cangle%20CAB)
d) Ta có:
AM xuất phát từ đỉnh A và vuông góc với BC