In triangle ABC, the line perpendicular to BC and passing though A intersects line BC in point D. Show that the area of triangle ABC is equal to 1/2 * AD * BC.
I know that the area of a triangle is 1/2 * base * height but I don't know how to prove/show it in this question. Anyone?
Notice that since AD is perp to BC , we have two right triangles - ADB and ACD
The area of ADB = (1/2) product of the legs = (1/2) (AD) (BD)
Similarly.....the area of ADC = (1/2) (AD) (DC)
So.....the combined areas = (1/2) (AD) ( BD + DC) =
(1/2) (AD) (BC) = area of triangle ABC