Prove that if two triangles are similar, their heights are in the same proportion as their side.

For a triangle ABC, the height of any side is a line
perpendicular to it passing through the third point. For example if AD is the height of
BC, AD is perpendicular to BC.

As AD is perpendicular to
BC, we have a right triangle ADC and we can write AD = AC* sin
C

Similarly the heights of the other sides can also be
written in terms of one of the other sides and the sine of the angle between
them.

Take two similar triangles ABC and A'B'C'. Let AD be
the height of BC and A'D' be the height of B'C'. As explained earlier, we can write AD =
AC*sin C and A'D' = A'C'*sin C'.

Similar triangles have the
same angles between adjacent sides. or the angle C between BC and AC is the same as the
angle C' between B'C' and A'C'

AD/A'D' = AC*sin C /
A'C'*sin C'

=> AD/A'D' =
AC/A'C'

This proves that the proportion of the height is
the same as the proportion of the corresponding sides. This is applicable to the heights
of all the sides. They have the same proportion as that of the corresponding
sides.

This proves that the heights of
similar triangles are in the same proportion as their
sides.