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Answer
In $\triangle A B C$ point $P$ and $D$ are on side $A C$, hence $B$ is common vertex of $\triangle ABD , \triangle BDC , \triangle ABC$ and $\triangle APB$ and their sides $AD , DC , AC$ and $AP$ are collinear. Heights of all the triangles are equal. Hence, areas of these triangles are proportinal to their bases. $AC =16, DC =9$
$\therefore A D=16-9=7$
$\therefore \frac{ A (\triangle ABD )}{ A (\triangle ABC )}=\frac{ AD }{ AC }=\frac{7}{16} \ldots \ldots$. triangles having equal heights
$\frac{ A (\triangle BDC )}{ A (\triangle ABC )}=\frac{ DC }{ AC }=\frac{9}{16} \ldots \ldots$. triangles having equal heights
$\frac{ A (\Delta ABD )}{ A (\Delta BDC )}=\frac{ AD }{ DC }=\frac{7}{9} \ldots \ldots$. triangles having equal heights
$\therefore A D=16-9=7$
$\therefore \frac{ A (\triangle ABD )}{ A (\triangle ABC )}=\frac{ AD }{ AC }=\frac{7}{16} \ldots \ldots$. triangles having equal heights
$\frac{ A (\triangle BDC )}{ A (\triangle ABC )}=\frac{ DC }{ AC }=\frac{9}{16} \ldots \ldots$. triangles having equal heights
$\frac{ A (\Delta ABD )}{ A (\Delta BDC )}=\frac{ AD }{ DC }=\frac{7}{9} \ldots \ldots$. triangles having equal heights
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