Question

In ΔABC, AM = MC = BM. Prove that ∠ABC = 90°.

Answer 1

Given:

• ΔABC such that AM = MC = BM,
• We need to prove that ∠ABC = 90°.

Proof:

We are given that: 

• AM = MC (i.e., M is the midpoint of AC),
• BM = AM (i.e., BM = MC).

This suggests a few important geometric properties that we can use to prove that ∠ABC = 90°.

Step 1: Analyze the triangle

Since M is the midpoint of AC, we have AM = MC

Also, BM = AM, which means that triangle ABM is isosceles, with AB = BM and AM = MC.

Therefore, triangle ABM is isosceles with equal sides AB = BM. 

Step 2: Recognize the type of triangle

Because M is the midpoint of side AC and BM = AM, the triangle ABM must be an isosceles triangle. In addition, if BM = AM, we have symmetry and the angle ∠BAM must be equal to ∠ABM.

Step 3: Apply the property of the isosceles triangle

In an isosceles triangle, the base angles are equal.

So ∠BAM = ∠ABM

Since the sum of angles in a triangle is 180°, we can write the equation for the angles in triangle ABM:

∠BAM + ∠ABM + ∠AMB = 180°

Since ∠BAM = ∠ABM, we have 2 × ∠BAM + ∠ABM = 180°

Thus, the angle ∠AMB is ∠AMB = 180° – 2 × ∠BAM.

Step 4: Use the perpendicular bisector property

Given that M is the midpoint of AC, the line BM is a perpendicular bisector of AC.

Therefore, ∠AMB = 90°

Thus, we have 90° = 180° – 2 × ∠BAM

Solving for ∠BAM:

2 × ∠BAM = 90°

⇒ ∠BAM = 45°

Step 5: Conclude that ∠ABC = 90°

Since ∠BAM = 45° and ∠ABM = 45°, the total angle ∠ABC is ∠ABC = 45° + 45° = 90°.