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Hint: Here we apply the area of triangle theorem where median divides the triangle in two equal areas.

Complete step-by-step answer:

In $\vartriangle ABC$

AO is the median, (CD is bisected by AB at O)

As you know, the median divides the triangle in two equal areas.

So,$area\left( {\vartriangle AOC} \right) = area\left( {\vartriangle AOD} \right)$ …… (1)

Also, in $\vartriangle BCD$

BO is the median, (CD is bisected by AB at O)

So, $area\left( {\vartriangle BOC} \right) = area\left( {\vartriangle BOD} \right)$ …… (2)

Adding (1) and (2) equations

$area\left( {\vartriangle AOC} \right) + area\left( {\vartriangle BOC} \right) = area\left( {\vartriangle AOD} \right) = area\left( {\vartriangle BOD} \right)$

$ \Rightarrow area\left( {\vartriangle ABC} \right) = area\left( {\vartriangle ABD} \right)$

Note: Whenever you come to this type of problem you have to know about median divide the triangle in two equal areas. So for easy solving you know about theorems regarding the area of a triangle.

Complete step-by-step answer:

In $\vartriangle ABC$

AO is the median, (CD is bisected by AB at O)

As you know, the median divides the triangle in two equal areas.

So,$area\left( {\vartriangle AOC} \right) = area\left( {\vartriangle AOD} \right)$ …… (1)

Also, in $\vartriangle BCD$

BO is the median, (CD is bisected by AB at O)

So, $area\left( {\vartriangle BOC} \right) = area\left( {\vartriangle BOD} \right)$ …… (2)

Adding (1) and (2) equations

$area\left( {\vartriangle AOC} \right) + area\left( {\vartriangle BOC} \right) = area\left( {\vartriangle AOD} \right) = area\left( {\vartriangle BOD} \right)$

$ \Rightarrow area\left( {\vartriangle ABC} \right) = area\left( {\vartriangle ABD} \right)$

Note: Whenever you come to this type of problem you have to know about median divide the triangle in two equal areas. So for easy solving you know about theorems regarding the area of a triangle.