Question

Prove that median of quilteral triangle is perpendicular to 3rd side

Answer 1

Consider an equilateral triangle ABC where AD is the median drawn from A to BC. We have to prove that AD is perpendicular to BC.

Since AD is the median, BD = CD.

Consider triangles ABD and ACD.

BD = CD

Since ∆ABC is equilateral, AB = AC

And AD = AD (common side)

So we have ∆ABD is congruent to ∆ACD. Thus,

⟨BAD = ⟨CAD

But we know that ⟨BAC = 60° since ∆ABC is equilateral. Well,

⟨BAC = ⟨BAD + ⟨CAD = 60°

2 ⟨BAD = 60° [⟨BAD = ⟨CAD]

⟨BAD = ⟨CAD = 30°

So that,

⟨ADC = ⟨ADB = 180° - (⟨ABD + ⟨BAD) = 180° - (60° + 30°) = 90°.

This tells that AD is perpendicular to BC.

Hence Proved!