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Let $\triangle ABC$ with lengths $a = BC, b=AC$, and $c=AB$ given. Find $D$ on $\overline{BC}$ such that $CD = \alpha a$ and find $E$ on $\overline{AC}$ such that $CE = \beta b$. Let $ABC$ be the area of $\triangle ABC$ and let $CDE$ be the area of $\triangle CDE$. Then $\frac{CDE}{ABC} = \alpha \beta.$

Area problem

I will prove this below. I would like my proof checked please and I wonder if there is a simpler proof.

Drop a perpendicular from $D$ to $D'$ on $\overleftrightarrow{AC}$ Let $h=DD'$. Drop a perpendicular from $B$ to $B'$ on $\overleftrightarrow{AC}$ Let $k=BB'$.

Then $\triangle CD'D \sim \triangle CB'B$. Hence $\dfrac hk = \dfrac{DD'}{BB'}= \dfrac{CD}{CB}=\dfrac{\alpha a}{a} = \alpha$.

So $\dfrac{CDE}{ABC} = \dfrac{\frac 12 \beta bh}{\frac 12bk}=\alpha \beta$.

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Your solution is perfectly fine, but it can be made shorter. You can use the following formula $[\triangle ABC] = \frac 12 A\cdot B \cdot \sin \theta$ ,where $\theta$ is the angle between sides $A$ and $B$.

Note that the two triangles have a common angle and we can use it to our advantage. Hence:

$$\frac{[\triangle CDE]}{[\triangle ABC]} = \frac{\frac 12 CD \cdot CE \cdot \sin DCE}{\frac 12 CA \cdot CB \cdot \sin ACB}= \frac{\alpha a \cdot \beta b}{ab} = \alpha \beta$$

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  • $\begingroup$ It's my habit to avoid trigonometry as much as possible while doing geometry. Of course your proof is better, I just tend to try to be as elementary as possible. $\endgroup$ – steven gregory May 22 '18 at 19:42

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