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Question
Prove by vector method that the median to the base of an isosceles triangle is perpendicular to the base
Solution
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"`
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