Question

ABC is an equilateral triangle. X, Y, Z are point AB,BC, CA respectively such that AX= BY=CZ.prove that XYZ is also an equilateral triangle

Answer 1

$\huge{\boxed{\bold{Question}}}$

ABC is an equilateral triangle. X, Y, Z are point AB,BC, CA respectively such that AX= BY=CZ.prove that XYZ is also an equilateral triangle

$\huge{\boxed{\bold{Answer}}}$

What to do

here we have bance

L.H.S=R.H.S.

Let's solve.

$\implies \rm \: \frac{1}{3} x - 4 = x( \frac{1}{2} + \frac{x}{3} ) \\ \\ \implies \rm \: \frac{x}{3} - 4 = x - ( \frac{3 + 2x}{6} ) \\ \\ \implies \rm \: \frac{x - 12}{3} = x - ( \frac{2x + 3}{ 6} ) \\ \\ \implies \rm \: \frac{x - 12}{3} = x - ( \frac{2x + 3}{ 6} ) \\ \\ \implies \rm \: \frac{x - 12}{3} = \frac{6x - 2x - 3}{6} \\ \\ \implies \rm \: \frac{x - 12}{3} = \frac{4x - 3}{6}$

Cancelling 6 (R.H.S) by 3 from L.H.S

$\ \longmapsto \sf \: \frac{x - 12}{1} = \frac{4x - 3}{2} \\ \\ \longmapsto\sf2( x - 12) = 4x - 3 \\ \\ \longmapsto \sf \: - 24 + 3 = 4x - 2x \\ \\ \longmapsto \sf \: - 21 = 2x \\ \\ \longmapsto \sf \bold{x = \frac{ - 21}{2} }$

Let's Prove this answer

$\leadsto \tt \: \frac{ \frac{ - 21}{2} }{3} - 4 = \frac{ - 21}{2} - ( \frac{1}{2} + ( - ) \frac{ \frac{21}{2} }{3} \\ \\ \leadsto \tt \: \frac{ - 21}{6} - 4 = \frac{ - 21}{2} - ( \frac{1}{2} - \frac{21}{6} ) \\ \\ \leadsto \tt \frac{ - 7}{2} - 4 = \frac{21}{2} - ( \frac{1}{2} - \frac{7}{2} ) \\ \\ \leadsto \tt \: \frac{ - 7 - 8}{2} = \frac{ - 21}{2} ( \frac{ - 6}{2} ) \\ \\ \leadsto \tt \: - \frac{ 15}{2} = \frac{21}{2} - ( - 3) \\ \\ \leadsto \tt \: - \frac{15}{2} = \frac{ - 21 + 6}{2} = - \frac{15}{2}$

Therefore ,L.H.S=R.H.S

CHECKED