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Geometry Difficulty 3.2 AMC 10/12 Find the answer Canada

In the diagram, point $D$ is
on side $AB$ of $\triangle ABC$ and points $E$ and $F$ are on $BC$ .

For some positive real number $t$ ,
the area of $\triangle DBE$ is $t$ , the area of $\triangle DEF$ is $t$ , the area of $\triangle DFC$ is $2t$ , and the area of $\triangle DAC$ is $4t$ . If the area of $\triangle DEC$ is $63$ , what is the area of $\triangle ABC$ ?

Pick one

Solution

The area of $\triangle DEC$ is
equal to the sum of the areas of $\$\triangle$
DEF $and$ $\triangle$ DFC$, and
so $t+2t=63$ or $3t=63$ or $t=21$ . The area of $\triangle ABC$ is equal to the sum of the
areas of $\triangle ADC$ , $\triangle DBE$ , $\triangle DEF$ and $\triangle DFC$ , which is equal to $4t+t+t+2t=8t=8(21)=168$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.