Easy proof of the intercept theorem.
We can see that $\large{\frac{A_{\triangle{ACE}}}{A_{\triangle{CDE}}}=\frac{AC}{CD}}$ (same height). Similarly, $\large{\frac{A_{\triangle{BCD}}}{A_{\triangle{CDE}}}=\frac{BC}{CE}}$.
Next, $A_{\triangle{ADE}}=A_{\triangle{BDE}}$ (area of two triangles with the same base between parallel lines).
Hence, $A_{\triangle{ACE}}=A_{\triangle{BCD}}$ and $\frac{A_{\triangle{ACE}}}{A_{\triangle{CDE}}}=\frac{A_{\triangle{BCD}}}{A_{\triangle{CDE}}}$
By transitivity, $$\frac{AC}{CD}=\frac{BC}{CE}$$
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Maximilian Böhmermann
Updated on October 13, 2020Comments
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Maximilian Böhmermann about 3 years
Im looking for a simple proof of the Intercept-Theorem in the Euclidean Plane $\mathbb{R}^2$. I can use analytic and synthetic Proofs and Theorems but students should be able to understand it. I've found long proofs with constructions but without use of other theorems.
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Vasili about 5 yearsCan you use the similarity of triangles? It's an easy way to prove this theorem.
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Maximilian Böhmermann about 5 yearsI can use it. But then i have to prove it. The proofs for the similarity theorem that i have seen are using the intercept theorem.
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Vasili about 5 yearsIn that case, you probably need to use areas of triangles and additional construction. It's quite simple.
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Maximilian Böhmermann about 5 yearsCan you give me a link for a proof or a link for a proof of the similarity theorem without the intercept theorem. Is there something simpler than the proofs on Wikipedia?
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Maximilian Böhmermann about 5 yearsCan you please show why $ \frac{AC}{CD}= \frac{AB}{ED}$.
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Vasili about 5 years@MaximilianBöhmermann: draw a line thru $E$ parallel to $AC$. The line will intersect $AB$ at point $H$. Using what we already proved, $AB/AH=BC/CE$. But $ADEH$ is a parallelogram so $AH=DE$.