Isosceles Triangles [Including Inequalities]
ICSE Class 9 · Mathematics · Chapter 10
What this chapter covers
- Isosceles triangles and their parts
- Angles opposite equal sides are equal
- The converse: sides opposite equal angles are equal
- Using both results in problems
- Inequalities in a triangle
- The greater side has the greater angle opposite it
- The greater angle has the greater side opposite it
- The sum of any two sides is greater than the third side
- The perpendicular is the shortest line from a point to a line
Key points
- An isosceles triangle has at least two equal sides; the angle between them is the vertex angle and the other two are the base angles.
- Angles opposite equal sides are equal: if AB = AC, then ∠ C = ∠ B.
- Sides opposite equal angles are equal: if ∠ B = ∠ C, then AC = AB.
- For vertex angle v and base angle b: v + 2b = 180°; each base angle is acute.
- A triangle is equilateral exactly when all its angles are 60°; an isosceles triangle with any one angle of 60° is equilateral.
- If two sides are unequal, the greater angle is opposite the greater side; if two angles are unequal, the greater side is opposite the greater angle.
- The hypotenuse is the longest side of a right-angled triangle; the side opposite an obtuse angle is the longest side of that triangle.
- The sum of any two sides of a triangle is greater than the third side, and the difference of any two sides is less than the third: |a - c| < b < a + c.
- Three lengths form a triangle exactly when the longest is less than the sum of the other two.
- Of all straight lines from a point to a given line, the perpendicular is the shortest; its length is the distance of the point from the line.
- These results are used in problems without proof; give the reason for each step in a few words.
Questions answered
Define an isosceles triangle and name its base angles when PQ = PR.
An isosceles triangle is a triangle in which at least two sides are equal. With PQ = PR, the base is QR and the base angles are ∠ Q and ∠ R.
In △ ABC, AB = BC. A student writes "∠ A = ∠ B". Is this right? Give the correct statement with a reason.
No. The angles opposite the equal sides are equal. Opposite AB is ∠ C and opposite BC is ∠ A, so ∠ A = ∠ C.
The vertex angle of an isosceles triangle is 104°. Find each base angle.
(180° - 104°)/2 = 38°.
Explain why a base angle of an isosceles triangle can never be 90°.
Both base angles are equal. If one were 90°, the two together would already make 180°, leaving nothing for the vertex angle, which is impossible.
In △ LMN, ∠ L = 45° and ∠ M = 67.5°. Is the triangle isosceles? If so, name the equal sides.
∠ N = 180° - 45° - 67.5° = 67.5° = ∠ M. Sides opposite equal angles are equal, so LN = LM. Yes, it is isosceles with vertex L.
In △ ABC, AB = AC and ∠ A = 60°. Show that the triangle is equilateral.
Each base angle is (180° - 60°)/2 = 60°, so all three angles are 60°. Sides opposite equal angles are equal, so AB = BC = CA.
D is a point on side BC of △ ABC with AD = BD = DC. Find ∠ BAC.
Let ∠ B = x and ∠ C = y. From AD = BD, ∠ BAD = x; from AD = DC, ∠ DAC = y. The angle sum of △ ABC gives x + y + (x + y) = 180°, so ∠ BAC = x + y = 90°.
In △ XYZ, XY = 9 cm, YZ = 7 cm and XZ = 12 cm. Name the greatest and the smallest angle.
The greatest angle is opposite the longest side XZ, so it is ∠ Y. The smallest is opposite the shortest side YZ, so it is ∠ X.
Past board exam questions
ICSE 2019 sample paper · 3 marks
In △ABC, ∠ABC = 66°, ∠DAC = 38°. CE is perpendicular to AB and AD is perpendicular to BC (E on AB, D on BC), and AD and CE intersect at P. Prove that: CP > AP
Show answer
In △ABD, ∠BAD = 90° − 66° = 24°, so ∠BAC = 24° + 38° = 62°.
In △ABC, ∠ACB = 180° − 66° − 62° = 52°.
In △BEC, ∠BCE = 90° − 66° = 24°, so ∠ACP = ∠ACB − ∠BCE = 52° − 24° = 28°.
In △APC, ∠PAC = 38° > ∠ACP = 28°.
The side opposite the greater angle is greater, so CP > AP.
ICSE 2019 sample paper · 3 marks
△ABC is isosceles triangle such that AB = AC. D is a point on side AB such that BC = CD. Given ∠BAC = 28°. Find the value of ∠DCA.
Show answer
AB = AC ⇒ ∠ABC = ∠ACB = (180° - 28°)/(2) = 76°.
In △BCD, BC = CD ⇒ ∠CDB = ∠CBD = 76°, so ∠BCD = 180° − 76° − 76° = 28°.
∠DCA = ∠ACB − ∠BCD = 76° − 28° = 48°.