4.2: Similar Triangles (2024)

Example \(\PageIndex{1}\)

Determine if the triangles are similar, and if so, write the similarity statement:

Solution

\[\angle C = 180^{\circ} - (65^{\circ} + 45^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ} \nonumber\]

\[\angle D = 180^{\circ} - (65^{\circ} + 45^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ} \nonumber\]

Therefore both triangles have the same angles and \(\triangle ABC \sim \triangle EFD\).

Answer: \(\triangle ABC \sim \triangle EFD\).

Theorem \(\PageIndex{1}\)

Two triangles are similar if two angles of one equal two angles of the other \((AA = AA)\).

In Figure \(\PageIndex{2}\), \(\triangle ABC \sim \triangle DEF\) because \(\angle A = \angle D\) and \(\angle B = \angle E\).

Proof

\(\triangle C = 180^{\circ} - (\angle A + \angle B) = 180^{\circ} - (\angle D + \angle E) = \angle F\).

Example \(\PageIndex{2}\)

Determine which triangles are similar and write a similarity statement:

Solution

\(\angle A = \angle CDE\) because they are corresponding angles of parallel lines. \(\angle C = \angle C\) because of identity. Therefore \(\triangle ABC \sim \triangle DEC\) by \(AA = AA\).

Answer: \(\triangle ABC \sim \triangle DEC\).

Example \(\PageIndex{3}\)

Determine which triangles are similar and write a similarity statement:

Solution

\(\angle A=\angle A\) identity. \(\angle ACB = \angle ADC=90^{\circ}\). Therefore

Also \(\angle B = \angle B\), identity, \(\angle BDC = \angle BCA = 90^{\circ}\). Therefore

Answer: \(\triangle ABC \sim \triangle ACD \sim \triangle CBD\).

Theorem \(\PageIndex{2}\)

The corresponding sides of similar triangles are proportional. This means that if \(\triangle ABC \sim \triangle DEF\) then

\(\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}\).

That is, the first two letters of \(\triangle ABC\) are to the first two letters of \(\triangle DEF\) as the last two letters of \(\triangle ABC\) are to the last two letters of \(\triangle DEF\) as the first and last letters of \(\triangle ABC\) are to the first and last letters of \(\triangle DEF\).

Example \(\PageIndex{4}\)

Find \(x\):

Solution

\(\angle A = \angle D\) and \(\angle B = \angle E\) so \(\triangle ABC \sim \triangle DEF\). By Theorem \(\PageIndex{2}\),

\(\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}\).

We will ignore \(\dfrac{AB}{DE}\) here since we do not know and do not have to find either \(AB\) or \(DE\).

\[\begin{array} {rcl} {\dfrac{BC}{EF}} & = & {\dfrac{AC}{DF}} \\ {\dfrac{8}{x}} & = & {\dfrac{2}{3}} \\ {24} & = & {2x} \\ {12} & = & {x} \end{array}\]

Check:

Answer: \(x = 12\).

Example \(\PageIndex{5}\)

Find \(x\):

Solution

\(\angle A = \angle A, \angle ADE = \angle ABC\), so \(\triangle ADE \sim \triangle ABC\) by \(AA = AA\).

\(\dfrac{AD}{AB} = \dfrac{DE}{BC} = \dfrac{AE}{AC}\).

We ignore \(\dfrac{AD}{AB}\).

\[\begin{array} {rcl} {\dfrac{DE}{BC}} & = & {\dfrac{AE}{AC}} \\ {\dfrac{5}{15}} & = & {\dfrac{10}{10 + x}} \\ {5(10 + x)} & = & {15(10)} \\ {50 + 5x} & = & {150} \\ {5x} & = & {150 - 50} \\ {5x} & = & {100} \\ {x} & = & {20} \end{array}\]

Check:

Answer: \(x = 20\).

Example \(\PageIndex{6}\)

Find \(x\):

Solution

\(\angle A = \angle CDE\) because they are corresponding angles of parallel lines. \(\angle C = \angle C\) because of identity. Therefore \(\triangle ABC \sim \triangle DEC\) by \(AA = AA\).

\(\dfrac{AB}{DE} = \dfrac{BC}{EC} = \dfrac{AC}{DC}\)

We ignore \(\dfrac{BC}{EC}\):

\[\begin{array} {rcl} {\dfrac{AB}{DE}} & = & {\dfrac{AC}{DC}} \\ {\dfrac{x + 5}{4}} & = & {\dfrac{x + 3}{3}} \\ {(x + 5)(3)} & = & {(4)(x + 3)} \\ {3x + 15} & = & {4x + 12} \\ {15 - 12} & = & {4x - 3x} \\ {3} & = & {x} \end{array}\]

Check:

Answer: \(x = 3\).

Example \(\PageIndex{7}\)

Find \(x\):

Solution

\(\angle A = \angle A\), \(\angle ACB = \angle ADC = 90^{\circ}\), \(\triangle ABC \sim \triangle ACD\).

\[\begin{array} {rcl} {\dfrac{AB}{AC}} & = & {\dfrac{AC}{AD}} \\ {\dfrac{x + 12}{8}} & = & {\dfrac{8}{x}} \\ {(x + 12)(x)} & = & {(8)(8)} \\ {x^2 + 12x} & = & {64} \\ {x^2 + 12x - 64} & = & {0} \\ {(x - 4)(x + 16)} & = & {0} \\ {x = 4\ \ \ \ \ \ \ \ x} & = & {-16} \end{array}\]

We reject the answer \(x = -16\) because \(AD = x\) cannot be negative.

Check, \(x = 4\)

Answer: \(x = 4\).

Example \(\PageIndex{8}\)

A tree casts a shadow 12 feet long at the same time a 6 foot man casts a shadow 4 feet long. What is the height of the tree?

Solution

In the diagram \(AB\) and \(DE\) are parallel rays of the sun. Therefore \(\angle A = \angle D\) because they are corresponding angles of parallel lines with respect to the transversal \(AF\). Since also \(\angle C = \angle F = 90^{\circ}\), we have \(\triangle ABC \sim \triangle DEF\) by \(AA = AA\).

\[\begin{array} {rcl} {\dfrac{AC}{DF}} & = & {\dfrac{BC}{EF}} \\ {\dfrac{4}{12}} & = & {\dfrac{6}{x}} \\ {4x} & = & {72} \\ {x} & = & {18} \end{array}\]

Answer: \(x = 18\) feet.

Historical Note

Thales (c. 600 B.C.) used the proportionality of sides of similar triangles to measure the heights of the pyramids in Egypt. His method was much like the one we used in Example \(\PageIndex{8}\) to measure the height of trees.

In Figure \(\PageIndex{7}\), \(DE\) represent the height of the pyramid and \(CE\) is the length of its shadow. \(BC\) represents a vertical stick and \(AC\) is the length of its shadow. We have \(\triangle ABC \sim \triangle CDE\). Thales was able to measure directly the lengths \(AC, BC\), and \(CE\). Substituting these values in the proportion \(\dfrac{BC}{DE} = \dfrac{AC}{CE}\), he was able to find the height \(DE\).