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In $\triangle ABC$ points $D,E,F$ are on the sides $AB,BC,CA$, respectively, with $AD=DB$, $CE=3BE$ and $AF=2CF$. If the area of $\triangle ABC$ is $480 cm^2$, how do we find the area of $\triangle DEF$?

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I suggest the use of vectors – UrošSlovenija May 16 '13 at 16:28
Vector is not in my curriculum yet.I need something simpler. – rahul May 16 '13 at 16:32
up vote 0 down vote accepted

Hint: Calculate the area of $ADF, BDE, ECF$. Subtract their sum from $ABC$ to get $DEF$.

Further Hint: $\frac{\mbox{Area } ADF } { \mbox {Area } ABC } = \frac{ AD } { AB} \times \frac { AF} { AC} $.

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That is what I thought.Didn't help. – rahul May 16 '13 at 16:50
@rahul See the further hint. – Calvin Lin May 16 '13 at 16:55
Thanks for helping! – rahul May 16 '13 at 17:12
This is a rather old question,but can you please tell me how you obtain those ratios? – rah4927 Dec 13 '13 at 13:16
@rah4927 They have a common angle of $ABC$, and it follows by the ratio of the lengths. Alternatively, if you know the sine rule area formula, Area $ABC = AB \times AC \times \sin \angle ABC$, then the result follows directly. – Calvin Lin Dec 13 '13 at 19:29


Drop the perpendicular from $B$ to $AC$ and $D$ to $AC$, and use that to find the area of $\triangle DAF$.

Use same idea for others.

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Denote $$ \vec a = \vec{AB}, \qquad \vec b = \vec{AC} $$ Then $$ \vec c = -\vec a + \vec b $$ and $$ \vec{DE} = \frac12 \vec a + \frac14 (-\vec a + \vec b) = \frac14 \vec a + \frac14 \vec b $$ $$ \vec{DF} = -\frac12 a + \frac23 \vec b $$ Area of the $\triangle DEF$ can be calculated as $$ \frac12 |\vec{DE} \times \vec{DF}| = \frac12 |(\frac14 \vec a + \frac14 \vec b) \times (-\frac12 a + \frac23 \vec b)| = \frac{7}{48} |\vec a \times \vec b| $$ since $\vec a \times \vec a = \vec b \times \vec b = 0$ and $\vec a \times \vec b = \vec b \times \vec a$.

Area of $\triangle ABC$ is also $$ \frac12 |\vec{a} \times \vec{b}| = 480 $$ implying $$ |\vec{a} \times \vec{b}| = 960 $$ Area of $\triangle DEF$ is therefore $$ \frac{7}{48} \cdot 960 = 140 $$

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