Question

in the given figure triangle abc and triangle dbc are on the same base bc and ad intersects bc at o if Al is perpendicular to bc and dm is perpendicular to bc prove that area of triangle abc by area of triangle dbc is equal to ao by do​

Answer 1

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

Step-by-step explanation:

Triangle ABC and DBC are on the same base BC.AD and BC intersect at O. Prove that- area of triangle ABC/area of triangle DBC=AO/DO

solution:

given : Triangle ABC and DBC are on the same base BC

to prove : $\frac{ar(ABC)}{ar(DBC)} = \frac{AO}{DO}$

Construction : Draw AM perpendicular to BC from point A and DN perpendicular to BC from point AD

proof:

$In \bigtriangleup AMO and \bigtriangleup DNO\\\angle AOM = \angle DON (v.o.a)\\\angle AMO = \angle DNO (90)\\So \\ \bigtriangleup AMO \cong \bigtriangleup DNO$

by AA criterian

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

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

therefore,

$\frac{(AO)^2}{(DO)^2} = \frac{(AM)^2}{(DN)^2}$

OR

$\frac{AO}{DO} = \frac{AM}{DN}$

now , area of triangle ABC = $\frac{1}{2}\times BC\times AM$

area of traingle DBC = $\frac{1}{2}\times BC\times DN$

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

Hence proved .

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