Question

如图,△ABC,D是AB的中点,AC=12,BC=5,CD=13/2.求证:△ABC为直角三角形.

Answer 1

cosA=(AD^2+AC^2-CD^2)/2AD*AC
cosA=(AB^2+AC^2-BC^2)/2AB*AC
∵D是AB的中点
∴AD=AB/2
∴AB=2AD
∴(AD^2+AC^2-CD^2)/2AD*AC=((2AD)^2+AC^2-BC^2)/2(2AD)*AC
(AD^2+144-169/4) / 2AD*AC =(4AD^2+144-25)/4AD*AC
2(AD^2+144-169/4)=4AD^2+144-25
4(AD^2+144-169/4)=2(4AD^2+144-25)
4AD^2+576-169=8AD^2+288-50
4AD^2=169
AD^2=169/4
AD=13/2
AB=13
∵AB^2=169
AC^2=144
BC^2=25
AC^2+BC^2=144+25=169=AB^2
∴△ABC为直角三角形