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9. Similarity
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Integrated Math 2

9.04 Applications of similarity

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Concept summary

In addition to the three theorems for justifying similarity, AA, SAS and SSS, there is one more theorem that is specific to right triangles.

Hypotenuse-Leg similarity (HL \sim) theorem

If the ratio of the hypotenuse and leg of one right triangle is equal to the ratio of the hypotenuse and leg of another right triangle, then the two triangles are similar.

Two right triangles. The hypotenuse and horizontal sides of the two triangles are congruent.

When two triangles are known to be similar, we can determine information about their corresponding sides and angles. In particular, there are special properties of the two triangles formed by constructing the midsegment of the larger triangle, as well as the triangles formed by constructing the altitude of a right triangle.

Triangle midsegment theorem

The midsegment connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle, and the length of this midsegment is half the length of the third side.

Triangle A B C with midpoint D on A B, and midpoint E on A C. A segment parallel to the B C is drawn from D to E. B C has a length of x, and D E has a length of x divided by 2.
Side-splitter theorem

If a line intersects two sides of a triangle and is parallel to the third side of the triangle, then it divides those two sides proportionally.

A triangle. Triangle A B C with point D on A B, and point E on A C. A segment parallel to the B C is drawn from D to E.
Converse of the side-splitter theorem

If a line divides two sides of a triangle proportionally, then the line is parallel to the third side of the triangle.

Triangle A B C with point D on A B, and point E on A C. A segment parallel to the B C is drawn from D to E.
Right triangle similarity theorem

The altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to the original triangle and to each other.

A right triangle with the altitude to the hypotenuse drawn, forming two similar triangles.

The right triangle similarity theorem also has the following two theorems related to it:

Geometric mean (altitude) theorem

The altitude to the hypotenuse of a right triangle divides the hypotenuse into two segments. The length of the altitude is the geometric mean of the lengths of the two segments of the hypotenuse.

A right triangle with the altitude to the hypotenuse having a length of a. The altitude to the hypotenuse divides the hypotenuse into 2 smaller segments: one of length b, and the other of length c. The equation, a squared equals b c, is shown.
Geometric mean (leg) theorem

The altitude to the hypotenuse of a right triangle separates the hypotenuse so that the length of each leg of the triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to the leg.

A right triangle with a hypotenuse of length c, and the altitude to the hypotenuse drawn. The altitude divides the hypotenuse into 2 smaller segments, one with length b. The leg of the triangle adjacent to the segment of length b has a length of a. The equation, c divided by a equals a divided by b, is shown.

Worked examples

Example 1

Determine whether \overline{KM} \parallel \overline{JN}. Justify your answer.

Triangle J L N with point K on J L, and point M on L N. A segment is drawn from K to M. Segment J K has a length of 1, K L has a length of 1.6, L M has a length of 8, and M N has a length of 5.

Approach

We can use the converse of the side-splitter theorem to determine whether \overline{KM} \parallel \overline{JN}. If \dfrac{LK}{KJ}=\dfrac{LM}{MN}, then the sides \overline{KM} and \overline{JN} are parallel.

Solution

We have that \dfrac{LK}{KJ}=\dfrac{1.6}{1}=1.6 and \dfrac{LM}{MN}=\dfrac{8}{5}=1.6.

Therefore, \overline{KM} \parallel \overline{JN} based on the converse of the side-splitter theorem.

Example 2

Solve for CD.

Right triangle A B C with right angle C and a point D on A B. A segment is drawn from C to D perpendicular to A B. Segment A D has a length of 8, and D B has a length of 4.

Approach

We can use the first corollary of the right triangle similarity theorem to determine the length of \overline{CD}, or we can equate corresponding ratios.

Solution

Equating ratios, we have:

\displaystyle \frac{AD}{CD}\displaystyle =\displaystyle \frac{CD}{DB}
\displaystyle \frac{8}{CD}\displaystyle =\displaystyle \frac{CD}{4}
\displaystyle \left(CD\right)^2\displaystyle =\displaystyle 32
\displaystyle CD\displaystyle =\displaystyle \sqrt{32}
\displaystyle CD\displaystyle =\displaystyle 4\sqrt{2}

Reflection

Note that if we used the corollary directly, we would get CD = \sqrt{AD \cdot DB} which then leads to CD = \sqrt{32}=4\sqrt{2}

Outcomes

M2.N.Q.A.1.B

Use appropriate quantities in formulas, converting units as necessary.

M2.G.SRT.B.3

Use congruence and similarity criteria for triangles to solve problems and to justify relationships in geometric figures.

M2.MP1

Make sense of problems and persevere in solving them.

M2.MP2

Reason abstractly and quantitatively.

M2.MP3

Construct viable arguments and critique the reasoning of others.

M2.MP4

Model with mathematics.

M2.MP5

Use appropriate tools strategically.

M2.MP6

Attend to precision.

M2.MP7

Look for and make use of structure.

M2.MP8

Look for and express regularity in repeated reasoning.

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