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Inscribed Similar Triangles

Division of a right triangle into two similar triangles.

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Inscribed Similar Triangles

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 \begin{align*}\triangle ADB, m \angle A = 90^{\circ}\end{align*}ADB,mA=90 and \begin{align*}\overline{AC} \bot \overline{DB}\end{align*}AC¯¯¯¯¯¯¯¯DB¯¯¯¯¯¯¯¯:

So, \begin{align*}\triangle ADB \sim \triangle CDA \sim \triangle CAB\end{align*}ADBCDACAB:

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? 

Examples

Example 1

Find the value of \begin{align*}x\end{align*}x.

Set up a proportion.

\begin{align*}\frac{\text{shorter leg in} \ \triangle SVT}{\text{shorter leg in} \ \triangle RST} &= \frac{\text{hypotenuse in} \ \triangle SVT}{\text{hypotenuse in} \ \triangle RST}\\ \frac{4}{x} &= \frac{x}{20}\\ x^2 &= 80\\ x &= \sqrt{80} = 4 \sqrt{5}\end{align*}shorter leg in SVTshorter leg in RST4xx2x=hypotenuse in SVThypotenuse in RST=x20=80=80=45

Example 2

Now find the value of \begin{align*}y\end{align*}y in \begin{align*}\triangle RST\end{align*}RST above.

Use the Pythagorean Theorem.

\begin{align*}y^2 + \left( 4 \sqrt{5} \right )^2 &= 20^2\\ y^2 + 80 &= 400\\ y^2 &= 320\\ y &= \sqrt{320} = 8 \sqrt{5}\end{align*}y2+(45)2y2+80y2y=202=400=320=320=85

Example 3

Find the value of \begin{align*}x\end{align*}x.

Separate the triangles to find the corresponding sides.

Set up a proportion.

\begin{align*}\frac{\text{shorter leg in} \ \triangle EDG}{\text{shorter leg in} \ \triangle DFG} &= \frac{\text{hypotenuse in} \ \triangle EDG}{\text{hypotenuse in} \ \triangle DFG}\\ \frac{6}{x} &= \frac{10}{8}\\ 48 &= 10x\\ 4.8 &= x\end{align*}shorter leg in EDGshorter leg in DFG6x484.8=hypotenuse in EDGhypotenuse in DFG=108=10x=x

Example 4

Find the value of \begin{align*}x\end{align*}x.

Set up a proportion.

\begin{align*}\frac{shorter \ leg \ of \ smallest \ \triangle}{shorter \ leg \ of \ middle \ \triangle} &= \frac{longer \ leg \ of \ smallest \ \triangle}{longer \ leg \ of \ middle \ \triangle}\\ \frac{9}{x} &= \frac{x}{27}\\ x^2 &= 243\\ x &= \sqrt{243} = 9 \sqrt{3}\end{align*}shorter leg of smallest shorter leg of middle 9xx2x=longer leg of smallest longer leg of middle =x27=243=243=93

Example 5

Find the values of \begin{align*}x\end{align*}x and \begin{align*}y\end{align*}y.

Separate the triangles. Write a proportion for \begin{align*}x\end{align*}x.

\begin{align*}\frac{20}{x} &= \frac{x}{35}\\ x^2 &= 20 \cdot 35\\ x &= \sqrt{20 \cdot 35}\\ x &= 10 \sqrt{7}\end{align*}20xx2xx=x35=2035=2035=107

Set up a proportion for \begin{align*}y\end{align*}y. Or, now that you know the value of \begin{align*}x\end{align*}x you can use the Pythagorean Theorem to solve for \begin{align*}y\end{align*}y. Use the method you feel most comfortable with.

\begin{align*}\frac{15}{y} &= \frac{y}{35} && (10 \sqrt{7})^2 + y^2 = 35^2\\ y^2 &= 15 \cdot 35 && \qquad 700 + y^2 =1225\\ y &= \sqrt{15 \cdot 35} && \qquad \qquad \quad \ y = \sqrt{525} = 5 \sqrt{21}\\ y &= 5 \sqrt{21}\end{align*}15yy2yy=y35=1535=1535=521(107)2+y2=352700+y2=1225 y=525=521

Review

Fill in the blanks.

  1. \begin{align*}\triangle BAD \sim \triangle \underline{\;\;\;\;\;\;\;\;\;} \sim \triangle \underline{\;\;\;\;\;\;\;\;\;}\end{align*}BAD
  2. \begin{align*}\frac{BC}{?} = \frac{?}{CD}\end{align*}BC?=?CD
  3. \begin{align*}\frac{BC}{AB} = \frac{AB}{?}\end{align*}BCAB=AB?
  4. \begin{align*}\frac{?}{AD} = \frac{AD}{BD}\end{align*}?AD=ADBD

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

Use the diagram to answer questions 7-10.

  1. Write the similarity statement for the three triangles in the diagram.
  2. If \begin{align*}JM = 12\end{align*}JM=12 and \begin{align*}ML = 9\end{align*}ML=9, find \begin{align*}KM\end{align*}KM.
  3. Find \begin{align*}JK\end{align*}JK.
  4. Find \begin{align*}KL\end{align*}KL.

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

  1. Fill in the blanks of the proof for the Inscribed Similar Triangles Theorem.

Given: \begin{align*}\triangle ABD\end{align*}ABD with \begin{align*}\overline{AC} \perp \overline {DB}\end{align*}AC¯¯¯¯¯¯¯¯DB¯¯¯¯¯¯¯¯ and \begin{align*}\angle DAB\end{align*}DAB is a right angle.

Prove: \begin{align*}\triangle ABD \sim \triangle CBA \sim \triangle CAD\end{align*}ABDCBACAD

Statement Reason
1. 1. Given
2. \begin{align*}\angle DCA\end{align*}DCA and \begin{align*}\angle ACB\end{align*}ACB are right angles 2.
3. \begin{align*}\angle DAB \cong \angle DCA \cong \angle ACB\end{align*}DABDCAACB 3.
4. 4. Reflexive PoC
5. 5. AA Similarity Postulate
6. \begin{align*}B \cong \angle B\end{align*}BB 6.
7. \begin{align*}\triangle CBA \cong \triangle ABD\end{align*}CBAABD 7.
8. \begin{align*}\triangle CAD \cong \triangle CBA\end{align*}CADCBA 8.

Review (Answers)

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

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Vocabulary

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^{\circ} 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.

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