In adjoining figure, $A P ⊥ B C, A D ∥ B C$ , then find
$A (△ A B C) : A (△ B C D)$

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Solution

Area of $△ = \frac{1}{2} \times b \times h$
$A B C$ and $B C D$ are $△$ between sample parallels and having the same base $B C$
$A 1 :$
Area of $△ A B C = \frac{1}{2} \times B C \times A P$
$A 2 :$
Area of $△ B C D = \frac{1}{2} \times B C \times A P$
Now
$\frac{A 1}{A 2} = \frac{1}{1} \Rightarrow 1 : 1$

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