Triangles

Similar figures

Figures that share their shape, though they may differ in size. All circles are similar to one another, as are all squares and all equilateral triangles.

Are all similar figures congruent? Are all congruent figures similar?

Every pair of congruent figures is similar, since they share both shape and size. The converse fails: two circles of different radii are similar but not congruent, because similarity only requires the same shape.

Similar polygons

Two polygons with the same number of sides are similar when all their corresponding angles are equal and all their corresponding sides are in the same ratio. Both conditions must hold.

Why is equal angles alone, or proportional sides alone, not enough to make two quadrilaterals similar?

A square and a non-square rectangle have all angles equal but sides not in one ratio. A square and a rhombus have sides in one ratio but unequal angles. Neither pair is similar, so polygons need both conditions.

Scale factor

The common ratio of corresponding sides of two similar polygons, also called the Representative Fraction. Maps and building blueprints are drawn using a chosen scale factor.

Similar triangles

Triangles whose corresponding angles are equal and whose corresponding sides are in the same ratio, written △ABC ~ △DEF. The order of the letters shows which vertices correspond, so A matches D, B matches E and C matches F.

Equiangular triangles

Two triangles whose corresponding angles are all equal. Thales observed that in such triangles the ratio of any two corresponding sides is always the same.

Basic Proportionality Theorem (Thales Theorem)

If a line parallel to one side of a triangle meets the other two sides at distinct points, it divides them in the same ratio: in △ABC with DE ∥ BC, AD/DB = AE/EC.

How is the Basic Proportionality Theorem proved using areas?

Join BE and CD. Using heights from E and D, ar(ADE)/ar(BDE) = AD/DB and ar(ADE)/ar(DEC) = AE/EC. Triangles BDE and DEC share base DE and lie between the parallels DE and BC, so their areas are equal, and the two ratios must match.

Converse of the Basic Proportionality Theorem

A line that cuts two sides of a triangle in equal ratios must be parallel to the third side: in △ABC, if AD/DB = AE/EC then DE ∥ BC.

In △ABC with DE ∥ BC, how do you show that AD/AB = AE/AC?

Start from the theorem's AD/DB = AE/EC and invert both sides to get DB/AD = EC/AE. Add 1 to each side, which gives AB/AD = AC/AE, and invert again.

In trapezium ABCD with AB ∥ DC and EF ∥ AB, how do you show that AE/ED = BF/FC?

Join diagonal AC, meeting EF at G. Since EF ∥ DC, apply the theorem in △ADC to get AE/ED = AG/GC. Since GF ∥ AB, apply it in △CAB to get AG/GC = BF/FC. The two results together give AE/ED = BF/FC.

AAA similarity criterion

If the corresponding angles of two triangles are equal, their corresponding sides are in the same ratio, so the triangles are similar.

AA similarity criterion

If two angles of one triangle equal two angles of another, the triangles are similar. By the angle sum property the third angles must then be equal too, so this is the AAA criterion in shorter form.

SSS similarity criterion

If the sides of one triangle are proportional to the sides of another, their corresponding angles are equal and the triangles are similar.

SAS similarity criterion

If one angle of a triangle equals one angle of another and the sides including those angles are proportional, the triangles are similar.

RHS similarity criterion

Two right triangles are similar when the hypotenuse and one other side of the first are in the same ratio as the hypotenuse and a side of the second.

Why is one condition enough to prove two triangles similar, when polygons in general need two?

For triangles the two conditions imply each other: equal corresponding angles force proportional sides (AAA), and proportional sides force equal angles (SSS). So checking either one settles similarity.

In the chapter's example, AB = 3.8, BC = 6, CA = 3√3 and RQ = 7.6, QP = 12, PR = 6√3, with ∠A = 80° and ∠B = 60°. How do you find ∠P?

Each ratio AB/RQ, BC/QP and CA/PR equals 1/2, so △ABC ~ △RQP by SSS. That makes ∠P correspond to ∠C. By the angle sum property, ∠C = 180° − 80° − 60° = 40°, so ∠P = 40°.

If lines AC and BD cross at O with OA · OB = OC · OD, how do you show ∠A = ∠C and ∠B = ∠D?

Rewrite the product as OA/OC = OD/OB. The angles AOD and COB are vertically opposite, so they are equal. The two sides around equal angles are proportional, so △AOD ~ △COB by SAS, and corresponding angles are equal.

A 90 cm girl walks 1.2 m/s away from a lamp 3.6 m high. How do you find her shadow's length after 4 seconds?

She is 4.8 m from the post. The post and girl are both vertical and share the angle at the shadow's tip, so the two triangles are similar by AA. With shadow x, (4.8 + x)/x = 3.6/0.9 = 4, so 3x = 4.8 and x = 1.6 m.

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