Finding Similar Triangles in a Figure

Spot the pair before any ratio is written.

A line parallel to the base

Similar triangles often hide inside one figure. Here is a triangle with a line drawn across it, parallel to its base. Parallel lines run in the same direction and never meet. The line cuts off a smaller triangle at the top.

So the figure holds two triangles: the small one at the top, and the whole triangle, which contains it. They share the top corner.

A triangle with a line across it, parallel to the base. The small triangle at the top has a base of 4 grid units, and the whole triangle has a base of 8.

Why the two are similar

For triangles, equal angles are enough to make them similar, so compare the angles one pair at a time.

The top angle belongs to both triangles, so that pair is equal.

The left side of the triangle crosses both the parallel line and the base. Because the two lines point in the same direction, the left side crosses them at the same angle. So the bottom left angle of the small triangle is equal to the bottom left angle of the whole triangle. Angles in matching positions where one line crosses two parallel lines are called corresponding angles, and they are always equal. The same is true on the right side.

All three angles of the small triangle are equal to the three angles of the whole triangle, so the two triangles are similar.

Match the sides by their angles

Corresponding sides are the sides across from equal angles. The base of the small triangle is across from the top angle, and so is the base of the whole triangle, so the two bases correspond. On the grid they are 4 and 8, so the scale factor from the small triangle to the whole one is 8 ÷ 4 = 2.

Every other pair of corresponding sides has the same factor. The left side of the small triangle runs from the top corner halfway down the left side of the whole triangle, so the whole left side is 2 times as long, as the factor says.

Compare with the whole side, never with the part that is left over. The small triangle’s left side corresponds to the whole left side of the big triangle, not to the piece below the parallel line. Here the two pieces happen to be equal, but in the first application below they are 3 m and 2 m, and only 3 against the whole 5 gives the scale factor.

Bowtie triangles

Two lines that cross between a pair of parallel lines make a bowtie: a triangle above the crossing point and a triangle below it. These two triangles are similar as well.

At the crossing point, the angle in the top triangle and the angle in the bottom triangle are vertically opposite: they are across the crossing from each other. They are equal. A third angle at the crossing point sits next to both of them. Each of the two, together with that third angle, fills a straight line, which is a half turn, 180°. So both are 180° minus the same angle, and they are equal.

Each crossing line meets both parallel lines, making a Z shape. The two angles inside the Z are called alternate angles, and they are equal too. At the lower parallel line, the angle inside the Z is equal to the corresponding angle at the upper line, and that angle is vertically opposite the angle inside the Z at the upper line.

So the angles of the top triangle are equal to the angles of the bottom triangle, and the triangles are similar.

Two crossing lines between parallel lines. The top triangle has a base of 3 grid units and the bottom triangle a base of 6.

In a bowtie, the matching corners swap sides

In the bowtie the bottom triangle is the top one turned upside down, so corresponding corners are diagonally opposite. The top left corner of the top triangle matches the bottom right corner of the bottom triangle, since the two are at the ends of one crossing line, and alternate angles are equal.

The two bases are across from the equal angles at the crossing point, so they correspond. They are 3 and 6 grid units long, so the scale factor from the top triangle to the bottom one is 6 ÷ 3 = 2. The crossing point is also 1 grid unit below the top line and 2 grid units above the bottom line: the heights are in the same ratio, 1 : 2.

The usual mistakes

Subtracting the bases. 8 − 4 = 4 is how much longer the big base is. A scale factor divides one corresponding side by the other: 8 ÷ 4 = 2.

Matching sides by where they sit on the page. In the bowtie, the top triangle is upside down compared with the bottom one. Find the equal angles first, then match the sides across from them.

Worked example: A Horizontal Beam Across the Triangular End Wall of a Roof

Question The end wall of a roof is a triangle ABC. The base BC is 6 m long, the top A is 4 m above BC, and the sloping edges AB and AC are each 5 m long. A horizontal beam DE joins the two sloping edges, with D on AB and E on AC, so that DE is parallel to BC. The point D is 3 m from A. (a) How long is the beam DE? (b) How high is the beam above the base BC?

  1. 1.DE is parallel to BC, so angle ADE is equal to angle ABC, and angle AED is equal to angle ACB, because they are corresponding angles. The angle at A belongs to both triangles. Triangle ADE and triangle ABC have equal angles, so they are similar.

    ABCDE6 m3 mAC = AB = 5 m4 m?equal angles: triangle ADE is similar to triangle ABC
    ABCDE6 m3 mAC = AB = 5 m4 m?equal angles: triangle ADE is similar to triangle ABC
    DE is parallel to BC, so the corresponding angles are equal and triangle ADE is similar to triangle ABC.

  2. 2.AD corresponds to AB, so the scale factor from triangle ABC to triangle ADE is ADAB = 35.

    ABCDE6 m3 mAC = AB = 5 m4 m?equal angles: triangle ADE is similar to triangle ABCscale factor: AD over AB = 3/5
    ABCDE6 m3 mAC = AB = 5 m4 m?equal angles: triangle ADE is similar to triangle ABCscale factor: AD over AB = 3/5
    AD corresponds to AB, so the scale factor is ADAB = 35.

  3. 3.DE corresponds to BC. (a) The beam is DE = 35 × 6 = 3.6 m long.

    ABCDE6 m3 mAC = AB = 5 m4 m3.6 mequal angles: triangle ADE is similar to triangle ABCscale factor: AD over AB = 3/5DE = 3/5 × 6 = 3.6 m
    ABCDE6 m3 mAC = AB = 5 m4 m3.6 mequal angles: triangle ADE is similar to triangle ABCscale factor: AD over AB = 3/5DE = 3/5 × 6 = 3.6 m
    (a) DE corresponds to BC: DE = 35 × 6 = 3.6 m.

  4. 4.A height is a length, so the heights are in the same ratio. The height of triangle ADE, measured down from A to the beam, is 35 × 4 = 2.4 m.

    ABCDE6 m3 mAC = AB = 5 m4 m3.6 m2.4 mequal angles: triangle ADE is similar to triangle ABCscale factor: AD over AB = 3/5DE = 3/5 × 6 = 3.6 mheight of ADE = 3/5 × 4 = 2.4 m
    ABCDE6 m3 mAC = AB = 5 m4 m3.6 m2.4 mequal angles: triangle ADE is similar to triangle ABCscale factor: AD over AB = 3/5DE = 3/5 × 6 = 3.6 mheight of ADE = 3/5 × 4 = 2.4 m
    Heights are in the same ratio: the height of triangle ADE is 35 × 4 = 2.4 m.

  5. 5.(b) The beam is 4 − 2.4 = 1.6 m above BC. Check: DEBC = 3.66 = 35 and 2.44 = 35.

    ABCDE6 m3 mAC = AB = 5 m4 m3.6 m2.4 m1.6 mequal angles: triangle ADE is similar to triangle ABCscale factor: AD over AB = 3/5DE = 3/5 × 6 = 3.6 mheight of ADE = 3/5 × 4 = 2.4 mbeam above BC: 4 − 2.4 = 1.6 m
    ABCDE6 m3 mAC = AB = 5 m4 m3.6 m2.4 m1.6 mequal angles: triangle ADE is similar to triangle ABCscale factor: AD over AB = 3/5DE = 3/5 × 6 = 3.6 mheight of ADE = 3/5 × 4 = 2.4 mbeam above BC: 4 − 2.4 = 1.6 m
    (b) The beam is 4 − 2.4 = 1.6 m above BC.

Answer: (a) 3.6 m; (b) 1.6 m

Common mistakes

  • Using the ratio of the two parts of AB, ADDB = 32, as the scale factor. The similar triangles are ADE and ABC, so the sides that correspond are AD and the whole of AB, and the ratio is 35.
  • Giving 2.4 m as the answer to part (b). That is the height of the small triangle, measured down from A. The question asks for the height of the beam above BC, which is the rest of the 4 m.

More pythagoras and similar shapes problems, worked step by step →

Worked example: The Width of a River Found by Sighting a Tree on the Far Bank

Question Aini wants to find the width of a river without crossing it. A tree T stands at the edge of the far bank. She stands at A on the near bank, directly opposite T, walks 20 m along the bank to a stake at C, and then walks 5 m further along the bank to D. At D she turns through a right angle and walks away from the river until, at E, the stake C is exactly in line with the tree T. She measures DE as 12 m. (a) How wide is the river? (b) How far is E from the tree T in a straight line?

  1. 1.TA crosses the river at right angles to the bank, and DE leaves the bank at right angles, so angle TAC and angle EDC are both 90°. The lines TE and AD cross at C, so angle ACT and angle DCE are equal, because they are vertically opposite angles. Triangle TAC and triangle EDC have equal angles, so they are similar.

    riverACDET20 m5 m12 m?two equal angles: triangle TAC is similar to triangle EDC
    riverACDET20 m5 m12 m?two equal angles: triangle TAC is similar to triangle EDC
    There are right angles at A and at D, and the angles at C are vertically opposite, so triangle TAC is similar to triangle EDC.

  2. 2.AC corresponds to DC, so the scale factor from triangle EDC to triangle TAC is 20 ÷ 5 = 4.

    riverACDET20 m5 m12 m?two equal angles: triangle TAC is similar to triangle EDCscale factor: 20 divided by 5 = 4
    riverACDET20 m5 m12 m?two equal angles: triangle TAC is similar to triangle EDCscale factor: 20 divided by 5 = 4
    AC corresponds to DC, so the scale factor is 20 ÷ 5 = 4.

  3. 3.TA corresponds to ED. (a) The river is 12 × 4 = 48 m wide.

    riverACDET20 m5 m12 m48 mtwo equal angles: triangle TAC is similar to triangle EDCscale factor: 20 divided by 5 = 4the river: TA = 12 × 4 = 48 m
    riverACDET20 m5 m12 m48 mtwo equal angles: triangle TAC is similar to triangle EDCscale factor: 20 divided by 5 = 4the river: TA = 12 × 4 = 48 m
    (a) TA corresponds to ED, so the river is 12 × 4 = 48 m wide.

  4. 4.In the small triangle, CE is the hypotenuse: CE2 = 52 + 122 = 25 + 144 = 169, so CE = √169 = 13 m. CT corresponds to CE, so CT = 13 × 4 = 52 m.

    riverACDET20 m5 m12 m48 m52 m13 mtwo equal angles: triangle TAC is similar to triangle EDCscale factor: 20 divided by 5 = 4the river: TA = 12 × 4 = 48 mCE2= 52+ 122= 169, so CE =√169 = 13 mCT = 13 × 4 = 52 m
    riverACDET20 m5 m12 m48 m52 m13 mtwo equal angles: triangle TAC is similar to triangle EDCscale factor: 20 divided by 5 = 4the river: TA = 12 × 4 = 48 mCE2= 52+ 122= 169, so CE =√169 = 13 mCT = 13 × 4 = 52 m
    CE2 = 52 + 122 = 169, so CE = 13 m, and CT = 13 × 4 = 52 m.

  5. 5.(b) E, C and T are in one straight line, so ET = 13 + 52 = 65 m. Check: 202 + 482 = 400 + 2304 = 2704 = 522.

    riverACDET20 m5 m12 m48 m52 m13 mtwo equal angles: triangle TAC is similar to triangle EDCscale factor: 20 divided by 5 = 4the river: TA = 12 × 4 = 48 mCE2= 52+ 122= 169, so CE =√169 = 13 mCT = 13 × 4 = 52 mET = 13 + 52 = 65 m
    riverACDET20 m5 m12 m48 m52 m13 mtwo equal angles: triangle TAC is similar to triangle EDCscale factor: 20 divided by 5 = 4the river: TA = 12 × 4 = 48 mCE2= 52+ 122= 169, so CE =√169 = 13 mCT = 13 × 4 = 52 mET = 13 + 52 = 65 m
    (b) E, C and T are in one straight line, so ET = 13 + 52 = 65 m.

Answer: (a) 48 m; (b) 65 m

Common mistakes

  • Matching TA with DC because both are drawn the same way on the page. Corresponding sides are opposite equal angles: TA and ED are each opposite the angle at C, and AC and DC each join the right angle to C.
  • Using the whole walk, AD = 25 m, as a side of the large triangle. The large triangle is TAC, and its side along the bank is AC = 20 m. The last 5 m belongs to the small triangle.

More pythagoras and similar shapes problems, worked step by step →

Practice Finding Similar Triangles in a Figure in the app