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.

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**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?**

Guest Jul 15, 2021

edited by
Guest
Jul 15, 2021

edited by Guest Jul 15, 2021

edited by Guest Jul 15, 2021

edited by Guest Jul 15, 2021

edited by Guest Jul 15, 2021

edited by Guest Jul 15, 2021

edited by Guest Jul 15, 2021

#1**+1 **

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

CPhill Jul 15, 2021