Building Similar Triangles

In one sentence

Congruent triangles tell you “these two segments are equal”. Similar triangles tell you “these four segments are in proportion”:

△ A B C \sim △ A^{'} B^{'} C^{'} ⟹ \frac{A B}{A ^{'} B ^{'}} = \frac{B C}{B ^{'} C ^{'}} = \frac{C A}{C ^{'} A ^{'}}

So when you see a proportion or a product equation, look for a pair of similar triangles whose corresponding sides are exactly those four segments. A handy trick: rewrite $ab = c d$ as $\frac{a}{c} = \frac{d}{b}$ . The endpoints of the two segments on top make one triangle, and the two on the bottom make the other.

Nothing ready-made in the figure? Then build one. The four most common ways to build them are the four stops below: draw a parallel line, draw an altitude, borrow angles from a circle, and copy an angle.

Predict first

$D E ∥ B C$ , $D$ is on $A B$ , and $A D : D B = 2 : 3$ . What is the ratio of the areas of $△ A D E$ and $△ A B C$ ?

2 : 3

2 : 5

4 : 9

4 : 25

Choose an answer first; then you can see the result.

Stop 1 · Draw a parallel line (Vol. 4 · similar triangles)

In $△ A B C$ , $D$ is on $A B$ with $A D = 3$ and $D B = 2$ . Through $D$ draw $D E ∥ B C$ , meeting $A C$ at $E$ . Given $B C = 10$ , find $D E$ .

Example: the “A” shape and the “8” shape

Step 1 / 4

Parallel lines give equal angles
Since $D E ∥ B C$ , corresponding angles are equal: $∠ A D E = ∠ A B C$ . Together with the shared $∠ A$ , the two triangles are similar. The shape looks like the letter A.

$△ A D E \sim △ A B C$
The ratio of similarity
Find DE
Flip to the other side

From now on, whenever “the segment ratio is not inside one triangle”, draw a parallel line through the dividing point to build an “A” or an “8”.

Stop 2 · Draw an altitude (Vol. 4 · proportional segments)

In right triangle $A B C$ , $∠ C = 9 0^{\circ}$ , $C D$ is the altitude to the hypotenuse, $A D = 2$ and $D B = 8$ . Find $C D$ and $A C$ .

Example: the altitude to the hypotenuse makes three similar triangles

Step 1 / 4

One altitude, three right triangles
The altitude $C D$ splits the big right triangle into two small right triangles. There are three right triangles in the figure.

$A D = 2, D B = 8, C D = ?$
The left and right ones are similar
The square of the altitude
A small one and the whole are similar too

$C D^{2} = A D \cdot D B$ , $A C^{2} = A D \cdot A B$ , $B C^{2} = B D \cdot A B$ : together these are called the projection theorem (also known as the geometric mean relations). No need to memorize them: draw the altitude, spot the similar triangles, and they are yours.

Stop 3 · Borrow angles from a circle (Vol. 5 · power of a point)

From a point $P$ outside a circle, draw two secants: one meets the circle at $A$ and $B$ , the other at $C$ and $D$ . $P A = 4$ , $A B = 5$ , $P C = 3$ . Find $C D$ .

Example: two secants

Step 1 / 4

Four segments, one product equation
We want $P A \cdot P B = P C \cdot P D$ , that is, $\frac{P A}{P D} = \frac{P C}{P B}$ . The tops $P A$ and $P C$ make $△ P A C$ ; the bottoms $P D$ and $P B$ make $△ P D B$ .

$P A = 4, P B = 9, P C = 3, P D = ?$
The circle gives a pair of equal angles
Write the proportion, cross-multiply
The result, and the limiting case

Stop 4 · Copy an angle (Vol. 5 · Ptolemy's theorem)

In the cyclic quadrilateral $A B C D$ , prove that $A B \cdot C D + A D \cdot B C = A C \cdot B D$ .

Two products added together will not fit into one pair of similar triangles. So cut $B D$ into two pieces and build two pairs.

Example: a proof of Ptolemy's theorem

Step 1 / 4

Split the right-hand side
Take a point $E$ on $B D$ ; then $A C \cdot B D = A C \cdot B E + A C \cdot E D$ . We want the first term to equal $A B \cdot C D$ and the second to equal $A D \cdot B C$ .

$A B \cdot C D + A D \cdot B C = ? A C \cdot B D$
Copy an angle
The second pair turns up for free
Add them up

Practice

Exercise 1Basic

In $△ A B C$ , $D$ is on $A B$ , $E$ is on $A C$ , and $D E ∥ B C$ . $A D = 4$ , $D B = 6$ , $B C = 15$ . Find $D E$ .

Exercise 2Advanced

In right triangle $A B C$ , $∠ C = 9 0^{\circ}$ , $C D ⊥ A B$ at $D$ , $A D = 9$ and $D B = 16$ . Find $A C$ .

Exercise 3Challenge

In trapezoid $A B C D$ , $A D ∥ B C$ , $A D = 4$ , $B C = 6$ , and the diagonals $A C$ and $B D$ meet at $O$ . The area of $△ A O D$ is $8$ . Find the area of trapezoid $A B C D$ .

In one sentence

Stop 1 · Draw a parallel line (Vol. 4 · similar triangles)

Example: the “A” shape and the “8” shape

Stop 2 · Draw an altitude (Vol. 4 · proportional segments)

Example: the altitude to the hypotenuse makes three similar triangles

Stop 3 · Borrow angles from a circle (Vol. 5 · power of a point)

Example: two secants

Stop 4 · Copy an angle (Vol. 5 · Ptolemy's theorem)

Example: a proof of Ptolemy's theorem

Practice

Mistake Museum · Exhibits from this stop

Using a piece of a side instead of the whole side

Forgetting to square the area ratio

Stops on this line