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Let-ABC-be-a-triangle-with-AB-AC-and-BAC-30-Let-A-be-the-reflection-of-A-in-the-line-BC-B-be-the-reflection-of-B-in-the-line-CA-C-be-the-reflection-of-C-in-the-line-AB-Show-that-A-B-C




Question Number 23592 by Tinkutara last updated on 02/Nov/17
Let ABC be a triangle with AB = AC  and ∠BAC = 30°. Let A′ be the reflection  of A in the line BC; B′ be the reflection  of B in the line CA; C′ be the reflection  of C in the line AB. Show that A′, B′, C′  form the vertices of an equilateral  triangle.
LetABCbeatrianglewithAB=ACandBAC=30°.LetAbethereflectionofAinthelineBC;BbethereflectionofBinthelineCA;CbethereflectionofCinthelineAB.ShowthatA,B,Cformtheverticesofanequilateraltriangle.
Answered by ajfour last updated on 02/Nov/17
Commented by Tinkutara last updated on 06/Nov/17
Commented by Tinkutara last updated on 06/Nov/17
Commented by ajfour last updated on 02/Nov/17
Given 𝛉=15° , to prove 𝛗=30°
Givenθ=15°,toproveϕ=30°
Commented by ajfour last updated on 02/Nov/17
OB ′=BB ′−OB    =2acos θ−(a/2)sec θ  MB ′=OB ′cos θ      =2acos^2 θ−(a/2)=(a/2)(4cos^2 θ−1)  A′M=OM+OD+A ′D    A′D=AD=(a/2)cot θ , so  A′M=OB ′sin θ+(a/2)tan θ+(a/2)cot θ     =(2acos θ−(a/2)sec θ)sin θ               +(a/2)tan θ+(a/2)cot θ    =(a/2)(4sin θcos θ+cot θ)  tan φ=((MB ′)/(A′M))         =(((a/2)(4cos^2 θ−1))/((a/2)(4sin θcos θ+cot θ)))        =(((a/2)(2+2cos 2θ−1))/((a/2)(2sin 2θ+(√((1+cos 2θ)/(1−cos 2θ))) )))  given θ=15°  ⇒  2θ=30° , so  tan φ=((2+(√3)−1)/(1+(√((2+(√3))/(2−(√3)))))) =(((√3)+1)/(1+2+(√3)))            =(1/( (√3))) .  ⇒ 𝛗 =30° and by symmetry      A′C ′=A′B ′  Hence △A′B ′C ′  is equilateral.
OB=BBOB=2acosθa2secθMB=OBcosθ=2acos2θa2=a2(4cos2θ1)AM=OM+OD+ADAD=AD=a2cotθ,soAM=OBsinθ+a2tanθ+a2cotθ=(2acosθa2secθ)sinθ+a2tanθ+a2cotθ=a2(4sinθcosθ+cotθ)tanϕ=MBAM=a2(4cos2θ1)a2(4sinθcosθ+cotθ)=a2(2+2cos2θ1)a2(2sin2θ+1+cos2θ1cos2θ)givenθ=15°2θ=30°,sotanϕ=2+311+2+323=3+11+2+3=13.ϕ=30°andbysymmetryAC=ABHenceABCisequilateral.
Commented by Tinkutara last updated on 02/Nov/17
Why AA′ bisects ∠C′A′B′?
WhyAAbisectsCAB?
Commented by ajfour last updated on 02/Nov/17
honestly, i dont know.
honestly,idontknow.
Commented by Tinkutara last updated on 06/Nov/17
Commented by ajfour last updated on 06/Nov/17
using which  app. have you typed  and drawn all these ?
usingwhichapp.haveyoutypedanddrawnallthese?
Commented by Tinkutara last updated on 15/Nov/17
Sir, can you show that BC∥B′C′?  Because I want to know how you  assumed ∠AMC′=90°.
Sir,canyoushowthatBCBC?BecauseIwanttoknowhowyouassumedAMC=90°.
Commented by ajfour last updated on 15/Nov/17
let x axis be along BC with D  as origin. Since △ABC is  isosceles, ∠OBC = ∠OCB  further OB = OC          and BE =CF  so  2BE = 2CF  ⇒      BB ′ =CC ′ =l  (say)       Hence y_(B ′)  =y_(C ′) =lsin θ  hence   B ′C ′ // BC .  And AD ⊥ BC   ⇒   ∠AMC ′ = ∠AMB ′ = 90° .
letxaxisbealongBCwithDasorigin.SinceABCisisosceles,OBC=OCBfurtherOB=OCandBE=CFso2BE=2CFBB=CC=l(say)HenceyB=yC=lsinθhenceBC//BC.AndADBCAMC=AMB=90°.
Commented by Tinkutara last updated on 15/Nov/17
What is y_(B′) ?
WhatisyB?
Commented by ajfour last updated on 15/Nov/17
I said, lets take BC as x axis  with D as origin.  y_(B ′)  is y coordinate of B ′ then.
Isaid,letstakeBCasxaxiswithDasorigin.yBisycoordinateofBthen.
Commented by Tinkutara last updated on 15/Nov/17
Thank you very much Sir!
ThankyouverymuchSir!

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