# Determine t such that the triangles have same area.

I have three points $A=(2,3), B=(6,4)$ and $C=(6,6).$ Given $\vec{AB}=\vec v$ and $\vec{BC}={0 \choose 2}$. I have also that for every $t\in [0,1]$ there is a point $D$ given as $\vec{AD}=t\vec{v}.$

My question is determine $t$ such that the area of triangle $ADC$ equals area of the triangle $DBC$.

My suggestions is Can I say that $D$ is on line $AB$ dividing the area of $ABC$ in to two equal parts, namely $ADC$ and $DBC$? If this is true then why is it true?

Thanks a lot.

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Yes, $t=\frac12$. Hint: The triangles $ADC$ and $DBC$ have the same altitude.
$$\overrightarrow{AD}=t\overrightarrow{v}$$ means that $AD$ and $v$ are linearly dependent. Also we can say $D\in[AB]$ because of $t\in[0,1]$. For $ADC$ and $DBC$ have same areas $t$ must be $1/2$ and hence $D=(4,7/2)$.