6.3: Method of similar triangles

Last updated
Save as PDF

Page ID: 23614

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

The proof of the Pythagorean theorem given above uses the method of similar triangles. To apply this method, one has to search for pairs of similar triangles and then use the proportionality of corresponding sides and/or equalities of corresponding angles. Finding such pairs might be tricky at first.

Exercise $\PageIndex{1}$

Let $ABC$ be a nondegenerate triangle and the points $X, Y$, and $Z$ as on the diagram. Assume $\measuredangle CAY \equiv \measuredangle XBC$. Find four pairs of similar triangles with these six points as the vertexes and prove their similarity.

$截屏2021-02-08 下午1.39.11.png$

Hint

By the AA similarity condition (Theorem 6.1.2), $\triangle AYC \sim \triangle BXC$. Conclude that $\dfrac{YC}{AC} = \dfrac{XC}{BC}$. Apply the SAS similarity condition to show that $\triangle ABC \sim \triangle YXC$.

Similarly, apply AA and equality of vertical angles to prove that $\triangle AZX \sim \triangle BZY$ and use SAS to show that $\triangle ABZ \sim \triangle YXZ$.