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

Pythagoras theorem

2 Answers
P dilip_k
Apr 1, 2018

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#

CW
Apr 1, 2018

see explanation.

Explanation:

enter image source here
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)"#

Related questions
Impact of this question
2986 views around the world
You can reuse this answer
Creative Commons License