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:
\(AM^2+MC^2=AC^2\)
\(AM^2+17^2=34^2\)
\(AM^2=34^2-17^2=867\left(cm\right)\)
\(AM=\sqrt{867}cm\)
