Question
✓
Answer
We can re-draw the fig.1.15(as shown above) where we add DO
which will be height of $\triangle B C D$.
Now, $\frac{ A (\triangle ABC )}{ A (\triangle BCD )}=\frac{ AP }{ DO }$
(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)
$\begin{aligned}
\Rightarrow & \frac{A(\triangle A B C)}{A(\triangle B C D)}=\frac{A P}{D O} \\
\Rightarrow & \frac{A(\triangle A B C)}{A(\triangle B C D)}=\frac{1}{1}
\end{aligned}$
( $\because$ the distance between the two parallel lines is always equal $\Rightarrow A P=D O)$
$\Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle BCD )}=1: 1$
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