Triangle Proportionality Theorem
Press on the numbers to see the steps of the proof.
1
Draw $$\small\Delta\mathtt{ABC}$$. Then draw $$\mathtt{\overline{DE}}$$ inside the triangle so that it is parallel with $$\mathtt{\overline{BC}}$$.
2
Line segments $$\mathtt{AB}$$ and $$\mathtt{AC}$$ form transversals to the parallel lines $$\mathtt{DE}$$ and $$\mathtt{BC}$$.
3
Corresponding angles are congruent.
$$\small\Delta\mathtt{ABC}$$ is similar to $$\small\Delta\mathtt{ADE}$$ by AA Similarity.
4
$$\mathtt{\frac{\color{blue}{BA}}{\color{red}{DA}} = \frac{\color{purple}{CA}}{\color{orange}{EA}}}$$, so $$\mathtt{\frac{\color{blue}{a + b}}{\color{red}{b}} = \frac{\color{purple}{c + d}}{\color{orange}{d}}}$$.

Rewrite this as $$\mathtt{\frac{\color{red}{b}}{\color{red}{b}} + \frac{a}{\color{red}{b}} = \frac{\color{orange}{d}}{\color{orange}{d}} + \frac{c}{\color{orange}{d}}}$$.

5
The fractions that equal 1 cancel, so $$\mathtt{\frac{a}{\color{red}{b}} = \frac{c}{\color{orange}{d}}}$$.
The segment $$\mathtt{DE}$$ divides $$\mathtt{\overline{\color{blue}{AB}}}$$ and $$\mathtt{\overline{\color{purple}{AC}}}$$ proportionally.
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