Isosceles triangles and inequalities

Congruence gives us powerful facts about single triangles too. In this lesson we prove that equal sides always face equal angles, and then go further: a longer side always faces a larger angle, and no side can be as long as the other two put together. These results let us find angles in isosceles triangles quickly, order the sides of a triangle from its angles, and decide whether three lengths can make a triangle at all.

Isosceles triangles

A triangle with two equal sides is isosceles. The angle between the equal sides is the vertex angle; the other two are the base angles.

Theorem. Angles opposite equal sides of a triangle are equal.

Proof. In triangle $ABC$ , let $AB = AC$ . Draw the bisector of $\angle A$ , meeting $BC$ at $D$ . In $\triangle BAD$ and $\triangle CAD$ : $AB = AC$ (given), $\angle BAD = \angle CAD$ (bisector), and $AD$ is common. By SAS the triangles are congruent, so $\angle B = \angle C$ by CPCT.

Converse. Sides opposite equal angles of a triangle are equal.

This can be proved with ASA in the same way. Together they tell us that a triangle with all three sides equal (equilateral) has all three angles equal, so each is $\displaystyle \frac{180°}{3} = 60°$ .

Example 1

The vertex angle of an isosceles triangle is $40°$ . Find the base angles.

The base angles are equal; call each $x$ . Then $40° + 2x = 180°$ , so $2x = 140°$ and $x = 70°$ .

Example 2

In triangle $PQR$ , $PQ = PR$ and the exterior angle at $Q$ , formed by extending $RQ$ beyond $Q$ , is $115°$ . Find all the angles of the triangle.

The interior angle at $Q$ is $180° - 115° = 65°$ . Since $PQ = PR$ , the angles opposite them are equal: $\angle R = \angle Q = 65°$ . Then $\angle P = 180° - 65° - 65° = 50°$ .

Example 3

In triangle $ABC$ , $AB = AC$ . The side $BA$ is extended beyond $A$ to $D$ so that $AD = AB$ . Prove that $\angle BCD = 90°$ .

Since $AB = AC$ , $\angle ABC = \angle ACB$ ; call this $x$ . Since $AD = AB = AC$ , triangle $ACD$ is also isosceles, so $\angle ACD = \angle ADC$ ; call this $y$ .

In triangle $BCD$ , the angles are $\angle B = x$ , $\angle D = y$ and $\angle BCD = x + y$ . Their sum is $180°$ :

$x + y + (x + y) = 180° \implies 2(x + y) = 180° \implies x + y = 90°$

So $\angle BCD = 90°$ .

Sides and angles: which is bigger?

Theorem. If two sides of a triangle are unequal, the angle opposite the longer side is larger.

Converse. In any triangle, the side opposite the larger angle is longer.

So the longest side faces the largest angle and the shortest side faces the smallest angle. In a right-angled triangle the right angle is the largest angle, so the hypotenuse is the longest side.

Example 4

In triangle $ABC$ , $\angle A = 70°$ and $\angle B = 50°$ . Arrange the sides in order of length.

$\angle C = 180° - 70° - 50° = 60°$ . The angles in order are $\angle B < \angle C < \angle A$ , so the sides opposite them are in the same order:

$AC < AB < BC$

( $AC$ is opposite $B$ , $AB$ is opposite $C$ , and $BC$ is opposite $A$ .)

The triangle inequality

Theorem. The sum of any two sides of a triangle is greater than the third side.

Think of walking from $B$ to $C$ : the straight path along $BC$ is shorter than going round through $A$ . So $AB + AC > BC$ , and the same holds for each pair of sides.

A useful consequence: each side is greater than the difference of the other two. So if two sides are $a$ and $b$ , with $a > b$ , the third side $c$ must satisfy

$a - b < c < a + b$

To test whether three lengths can form a triangle, it is enough to check that the two shorter ones add up to more than the longest one.

Example 5

Can a triangle have sides (a) $4$ cm, $7$ cm, $12$ cm (b) $5$ cm, $8$ cm, $11$ cm?

(a) $4 + 7 = 11 < 12$ . No triangle is possible.

(b) $5 + 8 = 13 > 11$ . A triangle is possible.

Example 6

Two sides of a triangle are $5$ cm and $9$ cm. Between which limits must the third side lie?

$9 - 5 < c < 9 + 5 \implies 4 < c < 14$

The third side must be more than $4$ cm and less than $14$ cm.

Practice

The vertex angle of an isosceles triangle is $100°$ . Find each base angle.
Each base angle of an isosceles triangle is $35°$ . Find the vertex angle.
One angle of an isosceles triangle is $50°$ , and it is a base angle. Find the other two angles.
In triangle $XYZ$ , $XY = XZ$ and $\angle X = (2a + 10)°$ , $\angle Y = (a + 25)°$ . Find all the angles.
In triangle $ABC$ , $\angle A = 45°$ and $\angle B = 75°$ . Which is the longest side and which is the shortest?
In triangle $PQR$ , $PQ = 6$ cm, $QR = 8$ cm and $RP = 5$ cm. Which is the largest angle and which is the smallest?
Can $3$ cm, $4$ cm and $8$ cm be the sides of a triangle? Can $6$ cm, $6$ cm and $11$ cm?
Two sides of a triangle are $7$ cm and $12$ cm. Find the range of the third side, and the least whole number of centimetres it can be.
In an equilateral triangle $ABC$ , the side $BC$ is extended beyond $C$ to $D$ . Find $\angle ACD$ .

Answers

Show answers

$\displaystyle \frac{180° - 100°}{2} = 40°$
$180° - 70° = 110°$
The other base angle is $50°$ and the vertex angle is $80°$ .
$\angle Y = \angle Z = (a + 25)°$ , so $(2a + 10) + 2(a + 25) = 180$ , giving $4a + 60 = 180$ and $a = 30$ . The angles are $\angle X = 70°$ , $\angle Y = \angle Z = 55°$ .
$\angle C = 60°$ . The longest side is opposite $75°$ : $AC$ . The shortest is opposite $45°$ : $BC$ .
The largest angle is opposite $QR$ : $\angle P$ . The smallest is opposite $RP$ : $\angle Q$ .
$3 + 4 = 7 < 8$ : no. $6 + 6 = 12 > 11$ : yes.
$5 < c < 19$ ; the least whole number is $6$ cm.
Each angle of the triangle is $60°$ , so $\angle ACD = 180° - 60° = 120°$ .

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Isosceles triangles and inequalities