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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?

 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
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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

 

 

cool cool cool

 Jul 15, 2021

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