Question

Prove that the trisector of all angles of any arbitrary triangle joins to form an equilateral triangle.

√ 50 Pts.
√ DETAILED SOLUTUON
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Answer 1

HEYA!!!!

Proof using trigonometric identities:

$sin3 0 = 4sin(60 + tetha) \sin(120 + tetha) - - - - - - (1)$
which by using the sum of two angles identity can be shown equal to

$\sin(30) = - 4 {sin}^{3} + 3 \sin$

points D,E, F are constructed on BC as shown clearly
$\alpha + \beta + \gamma$

Therefore the angles of triangle XEF are :
$\alpha (60 + \beta )and \: (60 + \gamma )$

from the figure,

$\sin(60 + \beta ) = \frac{DX}{XE} - - - - - - (2)$

and

$\sin(60 + \gamma ) = \frac{DX}{XF} - - - - - (3)$

$< ayc = 180 - \alpha - \gamma = 120 + \beta$

$< azb = 120 + \gamma$

The two sines applied to triangle AYC, AZB are :

$\sin(120 + \beta ) = \frac{AC}{AY} \sin( \gamma ) - - - - (5)$

and

$\sin(120 + \gamma ) = \frac{AB}{AZ} \sin( \beta ) - - - (6)$

Express the height of triangle ABC in two ways :

$h = AB \sin( 3\beta ) = AB.4 \sin( \beta ) \sin(60 + \beta ) \sin(120 + \beta )$

and

$h = AC \: \sin(3 \gamma ) = AC.4 \sin( \gamma ) \sin(60 + \gamma ) \sin(120 + \gamma )$

here, equation replaces as :

$h = 4AB \: \sin( \beta ) . \frac{DC}{XE} . \frac{AC }{AY} \sin( \gamma )$

and

$h = 4AC \: \sin( \gamma ) . \frac{dx}{xf} . \frac{AB}{AZ} \sin( \beta )$

sin numerators are equal :

$XE.AY = XF.AZ$

Since angle EXF and angle ZAY are equal and the sides forming these angles are in same ratio, triangle XEF and ZEY are equal.

similar angles AYZ and XFE equal 60+ a and similar angles AZY and XFE equal.

< AZY + < AZB + < BZX + < XZY = 360

$(60 + \beta ) + \: (120 + \gamma ) + (60 + \alpha ) + < zxy = 360$
Therefore, < ZXY = 60..

If we repeat to others we will get same angle..

Hence proved..

HOPE THIS HELPS U. @MP #MAHABOOBPASHA #☺☺

Answer 2

$\underline{\underline{\Huge\mathfrak{Answer ;}}}$

• See the attached file ⤴️