Question

In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that
$$\dfrac{Ar(\triangle ABC)}{Ar (\triangle DBC)} = \dfrac{AO}{DO}$$

Answer 1

$$Data: \space ABC\space and\space DBC\space are\space two\space triangles\space on\space the\space same\space base\space BC.\space If\space AD\space intersects\space BC\space at\space O.$$
$$To\space Prove:$$ $$\dfrac{Ar(\Delta ABC)}{Ar(\Delta DBC)} = \dfrac{AO}{DO}$$
$$Construction: \space Drawn$$ $$AP \perp BC$$ $$and$$ $$DM \perp BC.$$
$$Area\space of$$ $$\triangle = \dfrac{1}{2} \times base \times height$$
$$\dfrac{Ar(\Delta ABC)}{Ar (\Delta DBC)} = \dfrac{\dfrac{1}{2} \times BC \times AP}{\dfrac{1}{2} \times BC \times DM} = \dfrac{AP}{DM}$$
$$In$$ $$\triangle APO$$ $$and$$ $$\triangle DMO,$$
$$\angle APO = \angle DMO = 90^\circ$$ $$(construction)$$
$$\angle AOP = \angle DOM$$ $$(Vertically\space opposite\space angles)$$
$$\therefore \triangle APO \sim \triangle DMO$$
$$( Similarity\space criterion\space is\space AAA.)$$
$$\Rightarrow \dfrac{Ar(\Delta ABC)}{Ar(\Delta DBC)} = \dfrac{AO}{DO}$$