Skip to main content
K12 LibreTexts

7.11: Inscribed Similar Triangles

  • Page ID
    5815
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Division of a right triangle into similar triangles using an altitude.

    Inscribed Similar Triangles Theorem

    Remember that if two objects are similar, their corresponding angles are congruent and their sides are proportional in length. The altitude of a right triangle creates similar triangles.

    Inscribed Similar Triangles Theorem: If an altitude is drawn from the right angle of any right triangle, then the two triangles formed are similar to the original triangle and all three triangles are similar to each other.

    In \(\Delta ADB\), \(m\angle A=90^{\circ}\) and \(\overline{AC}\perp \overline{DB}\):

    f-d_8f94a81f6d6e18627a24bb14ed0a48e6ac658f490bb506661efa2f6a+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{1}\)

    So, \(\Delta ADB\sim \Delta CDA\sim \Delta CAB\):

    f-d_59b4e955d5eee8ab9073fb5e137b4abda1ecc9bb702a2196b47e1018+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{2}\)

    This means that all of the corresponding sides are proportional. You can use this fact to find missing lengths in right triangles.

    What if you drew a line from the right angle of a right triangle perpendicular to the side that is opposite that angle? How could you determine the length of that line?

    Example \(\PageIndex{1}\)

    Find the value of \(x\).

    f-d_d12f225661a2c00f86576c2273e012eb8561fdc1b393434047cac757+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{3}\)

    Solution

    Set up a proportion.

    \(\begin{aligned} \dfrac{\text{ shorter leg in }\Delta SVT}{\text{ shorter leg in }\Delta RST}&=\dfrac{\text{ hypotenuse in }\Delta SVT}{ \text{ hypotenuse in }\Delta RST} \\ \dfrac{4}{x}&=\dfrac{x}{20} \\ x^2&=80 \\ x&=\sqrt{80}=4\sqrt{5} \end{aligned}\)

    Example \(\PageIndex{2}\)

    Now find the value of \(y\) in \(\Delta RST\) above.

    Solution

    Use the Pythagorean Theorem.

    \(\begin{aligned} y^2+(4\sqrt{5})^2&=20^2 \\y^2+80&=400 \\ y^2&=320 \\ y&=\sqrt{320}=8\sqrt{5}\end{aligned}\)

    Example \(\PageIndex{3}\)

    Find the value of \(x\).

    f-d_2acb70c56937418b687c8657c7106c5c6267b8168cbf7e8b9e7991f2+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{4}\)

    Solution

    Separate the triangles to find the corresponding sides.

    f-d_54b7769afb1783c429453d3716496c24f4f8610049e8bf7c7568dfbb+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{5}\)

    Set up a proportion.

    \(\begin{aligned} \dfrac{\text{ shorter leg in } \Delta EDG}{\text{ shorter leg in } \Delta DFG}&=\dfrac{\text{ hypotenuse in }\Delta EDG}{\text{ hypotenuse in }\Delta DFG} \\ \dfrac{6}{x}&=\dfrac{10}{8} \\ 48&=10x \\ 4.8&=x \end{aligned}\)

    Example \(\PageIndex{4}\)

    Find the value of \(x\).

    f-d_239d4338244f87d20932f26cf26285ee377c3d51765c396884f36ef5+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{6}\)

    Solution

    Set up a proportion.

    \(\begin{aligned}\dfrac{\text{ shorter leg of smallest }\Delta}{\text{ shorter leg of middle } \Delta}=\dfrac{ \text{ longer leg of smallest }\Delta }{\text{ longer leg of middle }\Delta} \\ \dfrac{9}{x}&=\dfrac{x}{27} \\ x^2&=243 \\ x&=\sqrt{243}=9\sqrt{3}\end{aligned}\)

    Example \(\PageIndex{5}\)

    Find the values of \(x\) and \(y\).

    f-d_0cb387c3c521bae9dcac7a60aa53dadc08e59d31b735b70af2d1c5d9+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{7}\)

    Separate the triangles. Write a proportion for \(x\).

    Solution

    f-d_b6f2d6398038121a06ccacb83f202bd42595a8d774a6aa949f515b5d+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{8}\)

    \(\begin{aligned} \dfrac{20}{x}&=\dfrac{x}{35} \\ x^2&=20\cdot 35 \\ x&=\sqrt{20\cdot 35} \\ x&=10\sqrt{7}\end{aligned}\)

    Set up a proportion for y. Or, now that you know the value of x\) you can use the Pythagorean Theorem to solve for \(y\). Use the method you feel most comfortable with.

    \(\begin{array}{rlrl}
    \frac{15}{y} & =\frac{y}{35} & (10 \sqrt{7})^{2}+y^{2} & =35^{2} \\
    y^{2} & =15 \cdot 35 & 700+y^{2}&=1225 \\
    y & =\sqrt{15 \cdot 35} & y&=\sqrt{525}=5 \sqrt{21} \\& &
    y &=5 \sqrt{21}
    \end{array}\)

    Review

    Fill in the blanks.

    f-d_2a16e1e6d98896e8f9d88a72551fc8c9c6e6338e19febd47251a3b98+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{9}\)
    1. \(\Delta BAD\sim \Delta ______ \sim \Delta ______\)
    2. \(\dfrac{BC}{?}=\dfrac{?}{CD}\)
    3. \(\dfrac{BC}{AB}=\dfrac{AB}{?}\)
    4. \(\dfrac{?}{AD}=\dfrac{AD}{BD}\)

    Write the similarity statement for the right triangles in each diagram.

    1. f-d_cfef84eb13bbdfdb6ed9c403d93f987a791fc5f1782db8373e7b4164+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{10}\)
    2. f-d_82d536fd984657febfcbd62f5c2bc49357e301279d61a5f879e3dbe6+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{11}\)

    Use the diagram to answer questions 7-10.

    f-d_fdd82fa5d49de19b0bffde9f65050c1fd2c51e86e9c62cb5517c0cb6+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{12}\)
    1. Write the similarity statement for the three triangles in the diagram.
    2. If \(JM=12\) and \(ML=9\), find \(KM\).
    3. Find \(JK\).
    4. Find \(KL\).

    Find the length of the missing variable(s). Simplify all radicals.

    1. f-d_96ea0b1eeb54e0d687d61dcd83299376553c50edde5d93a17ac7bbfb+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{13}\)
    2. f-d_5ca2aefe8e9c10073191b5f24e12090efe6847dbcd339f3acf396f57+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{14}\)
    3. f-d_ef62b478837977549c1f746020c8108732cf71423d2bb94a9c8993f7+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{15}\)
    4. f-d_d72979bd42e7d3681862c66ea3b460bee01d47320ad99a698bcb581a+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{16}\)
    5. f-d_f9c86df9b8d84bf894883df8a12b77c99b7130b205eb39da3448d93e+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{17}\)
    6. f-d_165e624ea2a1671cd8d31911e47a782e5dfd9fee47ab30ac90e5e4ec+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{18}\)
    7. f-d_eb67a311015d121d3b0c3c804268b636c68bce4c766b8c63d5c374bc+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{19}\)
    8. f-d_db9a3eaa40a683422846e859a3c0158ba1d465b54efe015b21b33b02+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{20}\)
    9. f-d_2769805332ca338fe00448a0fc2d5de0c84a35a1adc36bcdf8c6d9b9+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{21}\)
    10. f-d_62d431f72047e42b700f265efeac567f1b8a2489103923db6f68f3e3+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{22}\)
    11. f-d_8c4a64125bdec9f8887736303881304d83a0dbacce0a217a06fbf26a+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{23}\)
    12. f-d_b7beb9e8b653d5b387a1f3c5dc80a8a847e26c1d8393f283ee37e44d+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{24}\)
    13. Fill in the blanks of the proof for the Inscribed Similar Triangles Theorem.
      f-d_2a16e1e6d98896e8f9d88a72551fc8c9c6e6338e19febd47251a3b98+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{25}\)

    Given: \(\Delta ABD\) with \(\overline{AC}\perp \overline{DB}\) and \(\angle DAB\) is a right angle.

    Prove: \(\Delta ABD\sim \Delta CBA\sim \Delta CAD\)

    Statement Reason
    1. 1. Given
    2. \(\angle DCA\) and \(\angle ACB\) are right angles 2.
    3. \(\angle DAB\cong \angle DCA\cong \angle ACB\) 3.
    4. 4. Reflexive PoC
    5. 5. AA Similarity Postulate
    6. \(B\cong \angle B\) 6.
    7. \(\Delta CBA\cong \Delta ABD\) 7.
    8. \(\Delta CAD\cong \Delta CBA\) 8.

    Review (Answers)

    To see the Review answers, open this PDF file and look for section 8.4.

    Vocabulary

    Term Definition
    Inscribed Similar Triangles Theorem The Inscribed Similar Triangles Theorem states that if an altitude is drawn from the right angle of any right triangle, then the two triangles formed are similar to the original triangle and all three triangles are similar to each other.
    Perpendicular Perpendicular lines are lines that intersect at a 90∘ angle. The product of the slopes of two perpendicular lines is -1.
    Proportion A proportion is an equation that shows two equivalent ratios.
    Pythagorean Theorem The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by \(a^2+b^2=c^2\), where a and b are legs of the triangle and c is the hypotenuse of the triangle.

    Additional Resources

    Video: Inscribed Similar Triangles Principles - Basic

    Activities: Inscribed Similar Triangles Discussion Questions

    Study Aids: Right Triangle Similarity Study Guide

    Practice: Inscribed Similar Triangles

    Real World: Inscribed Similar Triangles


    This page titled 7.11: Inscribed Similar Triangles is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform.

    CK-12 Foundation
    LICENSED UNDER
    CK-12 Foundation is licensed under CK-12 Curriculum Materials License
    • Was this article helpful?