Isosceles Triangles
ICSE · Class 9 · Mathematics
Flashcards for Isosceles Triangles — ICSE Class 9 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.
What is an isosceles triangle? Give its defining characteristic.
Answer
An isosceles triangle is a triangle in which at least two sides are equal to each other. The two equal sides are called the legs, and the third side is called the base.
What is an equilateral triangle and how does it relate to isosceles triangles?
Answer
An equilateral triangle is a triangle in which all three sides are equal. Since it has at least two equal sides, every equilateral triangle is also an isosceles triangle, but not every isosceles trian
State Theorem 1 for isosceles triangles.
Answer
Theorem 1: If two sides of a triangle are equal, then the angles opposite to them are also equal. In other words, in triangle ABC, if AB = AC, then ∠B = ∠C.
State Theorem 2 for isosceles triangles.
Answer
Theorem 2: If two angles of a triangle are equal, then the sides opposite to them are also equal. In other words, in triangle ABC, if ∠B = ∠C, then AB = AC.
What are the base angles of an isosceles triangle?
Answer
The base angles are the two angles that are opposite to the equal sides of an isosceles triangle. These angles are always equal to each other according to the properties of isosceles triangles.
What is the vertex angle of an isosceles triangle?
Answer
The vertex angle is the angle between the two equal sides of an isosceles triangle. It is the angle that is not a base angle and may have a different measure than the base angles.
In triangle ABC, AB = AC = 5 cm and ∠A = 40°. Find ∠B and ∠C.
Answer
Since AB = AC, triangle ABC is isosceles. By Theorem 1, ∠B = ∠C. Using angle sum property: ∠A + ∠B + ∠C = 180°. So 40° + ∠B + ∠B = 180°, which gives 2∠B = 140°, therefore ∠B = ∠C = 70°.
Prove that the bisector of the vertex angle of an isosceles triangle bisects the base at right angles.
Answer
Given: Triangle ABC with AB = AC, AD bisects ∠A. To prove: AD ⊥ BC and BD = DC. Proof: In triangles ABD and ACD, AB = AC (given), ∠BAD = ∠CAD (AD bisects ∠A), AD = AD (common). So △ABD ≅ △ACD by SAS.
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