Nguyên Hạo,"The median AM is also the altitude" is the property of \(\Delta ABC\) isosceles at A,so we can't deduce that \(\Delta ABC\) is the equilateral triangle.The question must give the length of BC

a) \(\Delta ABC\left(AB=AC\right)\)

AM is the median of ABC 

So, AM is also the the height of ABC.

=> \(AM\perp BC\)

b) ​ ​​ ​​ ​\(\Delta ABC\left(AB=AC\right)\)

AM is the median and the height of ABC

=> ABC is a equilateral triangle.

=> BC = 34 cm 

=> BM = MC = CB/2 = 34/2 = 17 cm (AM is the median)

b) \(\Delta AMC\left(\widehat{M}=90^o\right)\)

Pythagora's theorem:

\(AM^2+MC^2=AC^2\)

\(AM^2+17^2=34^2\)

\(AM^2=34^2-17^2=867\left(cm\right)\)

\(AM=\sqrt{867}cm\)