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.