Question

Prove by vector method that the median to the base of an isosceles triangle is perpendicular to the base

Answer 1

In isosceles ΔABC

Let AB = AC and AD is the median

D is the midpoint of BC

`bar"AD" = 1/2(bar"AB" + bar"AC")`

`bar"BC" = bar"BA" + bar"AC"`

`bar"DA" * bar"DB" = bar"AD" * ((-1)/2 bar"CB")`

= `- bar"AD"* (1/2 bar"BC")`

= `1/2 bar"AD" * bar"BC"`

= `1/4 (bar"AB" + bar"AC")*(bar"BA" + bar"AC")`

= `1/4(bar"AB" + bar"AC")*(bar"AC" - bar"AB")`

= `1/4[(bar"AC" * bar"AC") - (bar"AB" * bar"AB")]`

= `1/4("AC"^2 - "AB"^2)`

= `1/4(0)`

= 0

`bar"DA" * bar"DB"` = 0

 `bar"DA" ⊥ bar"DB"`