Similar Triangles: Examples and Problems with Solutions

If two angles of one triangle are congruent to two angles of another triangle, the triangles are similar.

Example 1:
Let \(\triangle ABC\) be a triangle and \(\overline{A'C'}\) parallel to \(\overline{AC}\). What can you say about \(\triangle ABC\) and \(\triangle A'BC'\)?

Solution:
Since \(\overline{A'C'} \parallel \overline{AC}\):
\(\angle BA'C' \cong \angle BAC\) (corresponding angles)
\(\angle BC'A' \cong \angle BCA\) (corresponding angles)
By AA similarity, \(\triangle ABC \sim \triangle A'BC'\).

If corresponding sides of two triangles are proportional, the triangles are similar.

Example 2:
Given vertices \(A(-2, 0)\), \(B(0, 4)\), \(C(2, 0)\) and \(P(-1, 1)\), \(Q(0, 3)\), \(R(1, 1)\), show that \(\triangle ABC \sim \triangle PQR\).

Solution:
Calculate side lengths:
\(AB = \sqrt{(0 + 2)^2 + (4 - 0)^2} = \sqrt{4 + 16} = 2\sqrt{5}\)
\(BC = \sqrt{(2 - 0)^2 + (0 - 4)^2} = \sqrt{4 + 16} = 2\sqrt{5}\)
\(CA = \sqrt{(-2 - 2)^2 + (0 - 0)^2} = 4\)
\(PQ = \sqrt{(0 + 1)^2 + (3 - 1)^2} = \sqrt{1 + 4} = \sqrt{5}\)
\(QR = \sqrt{(1 - 0)^2 + (1 - 3)^2} = \sqrt{1 + 4} = \sqrt{5}\)
\(RP = \sqrt{(-1 - 1)^2 + (1 - 1)^2} = 2\)
Ratios: \(\dfrac{AB}{PQ} = \dfrac{2\sqrt{5}}{\sqrt{5}} = 2\), \(\dfrac{BC}{QR} = 2\), \(\dfrac{CA}{RP} = 2\)
Since \(\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{CA}{RP}\), by SSS similarity \(\triangle ABC \sim \triangle PQR\).

If an angle of one triangle is congruent to an angle of another triangle, and the including sides are proportional, the triangles are similar.

Example 3:
Show that \(\triangle ABC \sim \triangle A'BC'\).

Solution:
\(\angle ABC \cong \angle A'BC'\) (given)
Check including sides: \(\dfrac{BA}{BA'} = \dfrac{10}{4} = 2.5\), \(\dfrac{BC}{BC'} = \dfrac{5}{2} = 2.5\)
Since the ratios are equal and the included angles congruent, by SAS similarity \(\triangle ABC \sim \triangle A'BC'\).

Problem Statement: In the diagram, \(\overline{A'C'} \parallel \overline{AC}\). Find length \(y = BC'\) and \(x = A'A\).

Solution:
Separate triangles: \(\triangle ABC\) and \(\triangle A'BC'\).
Since \(\overline{A'C'} \parallel \overline{AC}\), triangles are similar (AA).
Set up proportions: \(\dfrac{30 + x}{30} = \dfrac{22}{14} = \dfrac{y + 15}{y}\)
Solve for \(x\): \(\dfrac{30 + x}{30} = \dfrac{22}{14} \implies 14(30 + x) = 660\)
\(420 + 14x = 660 \implies 14x = 240 \implies x \approx 17.1\)
Solve for \(y\): \(\dfrac{22}{14} = \dfrac{y + 15}{y} \implies 22y = 14y + 210\)
\(8y = 210 \implies y = 26.25\)

Problem Statement: A light source at \(L\) (height 2 m) shines through pole top \(P'\) to mountain top \(M'\). Pole height = 20 m. Distance mountain-pole = 1000 m, pole-laser = 10 m. Find mountain height \(h\).

Solution:
Triangles \(\triangle LPP'\) and \(\triangle LMM'\) are similar (AA).
Proportions: \(\dfrac{LM}{LP} = \dfrac{MM'}{PP'} \implies \dfrac{1000 + 10}{10} = \dfrac{h - 2}{20 - 2}\)
\(\dfrac{1010}{10} = \dfrac{h - 2}{18} \implies 101 = \dfrac{h - 2}{18}\)
\(h - 2 = 1818 \implies h = 1820 \, \text{meters}\).

Problem Statement: Given \(\triangle ABC \sim \triangle A'B'C'\) with ratio \(k\), find ratio \(\dfrac{BH}{B'H'}\) of their altitudes.

Solution:
Since triangles are similar, \(\angle B \cong \angle B'\) and \(\angle BAH \cong \angle B'A'H'\).
Right triangles \(\triangle BHA\) and \(\triangle B'H'A'\) are similar (AA).
Therefore: \(\dfrac{BH}{B'H'} = \dfrac{AB}{A'B'} = k\).

Problem Statement: Chords \(\overline{BA'}\) and \(\overline{AB'}\) intersect at \(C\). Find a relationship between segments \(AC, BC, B'C, A'C\).

Solution:
Join \(\overline{BA}\) and \(\overline{B'A'}\).
\(\angle ABA' \cong \angle AB'A'\) (subtend same arc)
\(\angle BAB' \cong \angle BA'B'\) (subtend same arc)
So \(\triangle ABC \sim \triangle A'B'C\) (AA).
Proportions: \(\dfrac{BC}{B'C} = \dfrac{CA}{CA'}\)
Cross multiply: \(BC \cdot CA' = B'C \cdot CA\).

Problem Statement: In right \(\triangle ABC\), \(\overline{AM} \perp \overline{BC}\). How many similar triangles are there?

Solution:
Three similar triangles:
1. \(\triangle ABC \sim \triangle MBA\) (AA: right angle and \(\angle B\))
2. \(\triangle ABC \sim \triangle MAC\) (AA: right angle and \(\angle C\))
3. Consequently, \(\triangle MBA \sim \triangle MAC\).