Question

In the figure, ABC and DBC are two triangles on the same base BC . if AD intersects BC at O , show that area of traingle ABC proporrional area of traingle DBC =AO proportional DO​

Answer 1

$\textbf {Question:}$

In the figure, ΔABC & ΔDBC are two triangles on the same base BC. If AD intersects BC at O, show that $\dfrac{ar( \triangle abc)}{ar( \triangle dbc)} = \dfrac{ao}{do}$.

$\textbf {Solution:}$

$\tt{Given}$: ΔABC & ΔDBC on the same base BC.

$\tt{To\:prove}$: $\dfrac{ar( \triangle abc)}{ar( \triangle dbc)} = \dfrac{ao}{do}$

$\tt{Construction}$: AM perpendicular to BC and DN perpendicular to BC

$\tt{Proof}$:

In ΔAMO & ΔDNO,

$\angle AOM = \angle DON$

$\angle AMO = \angle DNO$

$\therefore \triangle AMO \approx \triangle DNO$

$\implies \dfrac{AM} {DN} = \dfrac {AO} {DO} \: (CSST)$

Now,

$\dfrac{ar(ABC)} {ar(DBC)} = \dfrac { \frac{1}{2} \times BC \times AM} { \frac{1}{2} \times BC \times DN} \\ \\ \dfrac{ar(ABC)} {ar(DBC)} = \dfrac{AM} {DN} = \dfrac{AO} {DO} \\ \\ Hence, proved.$

Answer 2

Given :

∆ ABC and ∆DBC have same base BC.

To proof :

$\sf{\large{\frac{ar(\triangle ABC)}{ar(\triangle DBC)}=\frac{AO}{DO}}}$

Construction :

Draw AM perpendicular to BC from point A and draw DN perpendicular to BC from point D.

Proof :

In ∆ AMO and ∆ DNO,

$\sf{\angle AMO=\angle DNO}$ [right angles]

$\sf{\angle AOM=\angle DON}$ [vertically opposite angles]

So by AA criteria,

$\sf{\triangle AOM\approx\triangle DON}$

$\sf{\large{\frac{ar(\triangle AOM)}{ar(\triangle DON}=\frac{(AO)^2}{(DO)^2}=\frac{(AM)^2}{(DN)^2}}}$

[Ratio of the area of two similar triangles is equal to the square of their corresponding sides.]

⇒ $\sf{\frac{(AO)}{(DO)}=\frac{(AM)}{(DN)}}$

Now,

Area of ∆ ABC = $\sf{\frac{1}{2}\times BC\times AM}$

Area of ∆ DBC = $\sf{\frac{1}{2}\times BC\times DN}$

⇒ $\sf{\large{\frac{ar(\triangle ABC)}{ar(\triangle DBC)}=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times BC\times DN}=\frac{AM}{DN}=\frac{AO}{DO}}}$