Statement 1: If two sides of a triangle are equal, then the angles opposite the equal sides are equal.
Statement 2: If two angles of a triangle are equal, then the sides opposite the equal angles have equal lengths.
We have proved the first statement in an earlier grade. Is the second statement true? Can you prove it?
(Hint: Draw the altitude from the vertex containing the third angle.)
Step-by-step solution
To find: Prove that AB = AC (the sides opposite the equal angles)
Idea: Statement 2 is the converse of Statement 1. The altitude from A (the vertex with the third angle) cuts the triangle into two right triangles. They have two equal angles and a common side, so they are congruent, and the sides we want are matching sides.
- Let △ABC have ∠B = ∠C. The side opposite ∠B is AC and the side opposite ∠C is AB, so we must show AB = AC.½ mark
- Draw the altitude AD from A to BC. Then AD ⊥ BC, so ∠ADB = ∠ADC = 90°.½ mark
- Compare △ABD and △ACD:
∠ABD = ∠ACD (given),
∠ADB = ∠ADC (each 90°),
AD = AD (common side).1 mark - Two angles and a non-included side match, so △ABD ≅ △ACD by the AAS congruence rule.½ mark
- Matching parts of congruent triangles are equal, so AB = AC (CPCT). Statement 2 is true: Statement 1 and its converse are both true.½ mark
Check: Draw BC = 6 cm with angles of 50° at B and at C. Measuring gives AB = AC ≈ 4.67 cm (each is 3 ÷ cos 50°).
Answer to write in the exam
Given: In △ABC, ∠B = ∠C. To prove: AB = AC.
Construction: Draw AD ⊥ BC.
In △ABD and △ACD:
∠ABD = ∠ACD (given)
∠ADB = ∠ADC = 90° (AD ⊥ BC)
AD = AD (common)
∴ △ABD ≅ △ACD (AAS rule)
∴ AB = AC (CPCT)
∴ Yes, Statement 2 is true.
Common mistakes that cost marks
- Using SAS or SSS. At the start no two sides are known to be equal, so the rule must use two angles and a side: AAS.
- Assuming D is the midpoint of BC. BD = DC follows only after the triangles are proved congruent; it cannot be used to prove it.
- Proving Statement 1 again by mistake. Here the angles are given and the sides must be proved equal.
How this can come in the exam
In △PQR, ∠Q = ∠R = 65° and PQ = 7 cm. Find PR and ∠P.
Show answer
Sides opposite equal angles are equal: PR (opposite ∠Q) = PQ (opposite ∠R) = 7 cm (1 mark).∠P = 180° − 65° − 65° = 50° (1 mark).
In △XYZ, ∠X = ∠Z. Which two sides must be equal?
- XY and YZ
- XY and XZ
- XZ and YZ
- No two sides need be equal
Show answer
(A) XY and YZ
The side opposite ∠X is YZ and the side opposite ∠Z is XY, so XY = YZ.
Try one yourself
In △ABC, ∠A = ∠C = 70° and AB = 5 cm. Find BC and ∠B.
Show answer
BC is opposite ∠A and AB is opposite ∠C, so BC = AB = 5 cm. ∠B = 180° − 140° = 40°.
More questions like this
- Proposition P: If it rains, then the road is wet.
Converse Q: If the road is wet, then it has rained. - Proposition P: If a number is a multiple of 6, then it is a multiple of 3.
Converse Q: If a number is a multiple of 3, then it is a multiple of 6. - In this example, n is any positive integer.
Proposition P: If n is a perfect square, then it has an odd number of factors.
Converse Q: If n has an odd number of factors, then it is a perfect square.
You may recall that we came across these statements in the previous grade. Which of them are true? - Consider the following argument.
Each factor of a number has a ‘partner’ factor such that their product yields the given number, e.g., 5 is a factor of 35, and 5 × 7 = 35. Here, 7 is the partner factor of 5, and vice-versa. Let us focus on factor-partner pairs of numbers, e.g., (1, 12), (2, 6), (3, 4) are the pairs for the number 12.
If a number has an odd number of factors, there must be a factor-partner pair in which the same number repeats (e.g., the partner factor of 5 in 25). If not, the given number will have an even number of factors since each factor can be paired with its factor pair.
Thus, if a number has an odd number of factors, then it is a perfect square.
What does this argument prove? Statement P or Q? - This only proves Q. It doesn’t prove that every square number has an odd number of factors (Proposition P).
Is P true?