Q4 of 49 Page 5

In adjoining figure, AP ⊥ BC, AD || BC, then find A(ΔABC) : A (ΔBCD)


We can re-draw the fig.1.15(as shown above) where we add DO


which will be height of ΔBCD.


Now,


(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)


⇒


⇒


(∵ the distance between the two parallel lines is always equal ⇒ AP = DO)


⇒ = 1:1


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