Question

If LMN is an equilateral triangle and X is the mid point of LN. prove that `MX^2 = 3/4MN^2`?

Pythagoras theorem

Answer 1

In equilateral `DeltaMLN`,`ML=LN=NM`.

Given `X` is the mid point of `LN` `LX=XN`

We have for `DeltasMLXandMNX`

`ML=MN,LX=XNandMX "common"`

`DeltasMLXandMNX` are congruent.

So `angleMXN=angleMXL=90^@`

By Pythagoras theorem

`MX^2=MN^2-NX^2`

`=>MX^2=MN^2-(1/2LN)^2`

`=>MX^2=MN^2-(1/2MN)^2`

`=>MX^2=MN^2-1/4MN^2`

`=>MX^2=3/4MN^2`

Answer 2

see explanation.

Explanation:

Given that `DeltaLMN` is equilateral,
`=> LM=MN=NL, and angleLMN=angleMNL=angleNLM=60^@`,
given `X` is the midpoint of `LN`,
`=> angleMXL=angleMXN=90^@`
In `DeltaMLX, MLsin60=MX ------ Eq(1)`
In `DeltaMNX, MNsin60=MX ------ Eq(2)`
`Eq(1)xxEq(2), => ML*MN*sin^2 60=MX^2`
`=> MX^2=MN^2*(sqrt3/2)^2`
`=> MX^2=3/4*MN^2 " (proved)"`