Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 16277 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17

$i n t r i a n g l e : A \overset{Δ}{B C}, c o n s t e a c t 3 e q u i l a t e r a l$ $t r i a n g l e o n e a c h s i d e s .$ $p r o v e t h a t : A L = B M = C N .$
	Answered by mrW1 last updated on 20/Jun/17

	$AB = NB$ $BL = BC$ $∠ ABL = ∠ B + 60 ° = ∠ NBC$ $\Rightarrow Δ ABL \equiv Δ NBC$ $\Rightarrow AL = NC$ $similarly$ $NC = MB$ $\Rightarrow AL = BM = CN$ $additionally :$ $∠ BCG = ∠ BLG$ $∠ CBG = ∠ CLG$ $\Rightarrow ∠ BCG + ∠ CBG = ∠ BLG + ∠ CLG = 60^{°}$ $\Rightarrow ∠ BGC = 180 ° - (∠ BCG + ∠ CBG) = 120 °$ $similarly$ $\Rightarrow ∠ AGB = 120 °$ $\Rightarrow ∠ CGA = 120 °$
	Commented by ajfour last updated on 20/Jun/17

	$w o n d e r f u l S i r!$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17

	$t h a n k y o u d e a r m r W 1 . s o n i c e .$
	Answered by ajfour last updated on 20/Jun/17
	$In ΔNAC CN^( 2) =b^2 +c^2 −2bccos (A+π/3) In ΔMCB BM^( 2) =a^2 +b^2 −2abcos (C+π/3) CN^( 2) −BM^( 2) = c^2 −a^2 −2b[ccos (A+π/3)−acos (C+π/3)] =c^2 −a^2 −2b{((ccos A)/2)−((c(√3)sin A)/2) −((acos C)/2)+((a(√3))/2)sin C } =c^2 −a^2 −b[ccos A−acos C] =c^2 −a^2 −b[((c(b^2 +c^2 −a^2 ))/(2bc))−((a(a^2 +b^2 −c^2 ))/(2ab))] =c^2 −a^2 −(1/2)[b^2 +c^2 −a^2 −a^2 −b^2 +c^2 ] = c^2 −a^2 −(c^2 −a^2 ) =0 Hence CN=BM similarly CN^( 2) −AL^2 =0 can be proved . ⇒ AL = BM = CN .$
	$I n Δ N A C$ ${C N}^{2} = b^{2} + c^{2} - 2 b c \cos (A + π / 3)$ $I n Δ M C B$ ${B M}^{2} = a^{2} + b^{2} - 2 a b \cos (C + π / 3)$ ${C N}^{2} - {B M}^{2} = c^{2} - a^{2}$ $- 2 b [c \cos (A + π / 3) - a \cos (C + π / 3)]$ $= c^{2} - a^{2} - 2 b {\frac{c \cos A}{2} - \frac{c \sqrt{3} \sin A}{2}$ $- \frac{a \cos C}{2} + \frac{a \sqrt{3}}{2} \sin C}$ $= c^{2} - a^{2} - b [c \cos A - a \cos C]$ $= c^{2} - a^{2} - b [\frac{c (b^{2} + c^{2} - a^{2})}{2 b c} - \frac{a (a^{2} + b^{2} - c^{2})}{2 a b}]$ $= c^{2} - a^{2} - \frac{1}{2} [b^{2} + c^{2} - a^{2} - a^{2} - b^{2} + c^{2}]$ $= c^{2} - a^{2} - (c^{2} - a^{2}) = 0$ $H e n c e C N = B M$ $s i m i l a r l y {C N}^{2} - {A L}^{2} = 0$ $c a n b e p r o v e d .$ $\Rightarrow A L = B M = C N .$
	Commented by ajfour last updated on 20/Jun/17

	$v e r y f o o l i s h . .$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17

	$t h a n k y o u d e a r m r A j f o u r . i t i s b e a u t i f u l .$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17

	$A (a), B (b), C (c), M (m), N (n), L (l)$ $ω^{2} + ω + 1 = 0, \frac{1}{ω} = - (1 + ω), \frac{1}{ω^{2}} = ω$ $ω = \frac{- 1 + i \sqrt{3}}{2}$ $c + ω . l + ω^{2} . b = 0 \Rightarrow l = - \frac{c + ω^{2} . b}{ω} = (1 + ω) (c + ω^{2} . b) =$ $= c + ω^{2} . b + ω . c + ω^{3} . b = b + c + ω . c + ω^{2} . b$ $b + ω . n + ω^{2} . a = 0 \Rightarrow n = - \frac{b + ω^{2} . a}{ω} = (1 + ω) (b + ω^{2} . a) =$ $= a + b + ω . b + ω^{2} . a$ $a + ω . m + ω^{2} . c = 0 \Rightarrow m = - \frac{a + ω^{2} . c}{ω} = (1 + ω) (a + ω^{2} . c) =$ $= a + c + ω . a + ω^{2} . c$ $A L = a - l = a - (b + c + ω . c + ω^{2} . b) = a - b - c - ω . c + (1 + ω) . b =$ $= a - b - c - ω . c + b + ω . b = (a - c) + ω (b - c) =$ $= a - c + \frac{- 1 + i \sqrt{3}}{2} (b - c) = \frac{1}{2} (2 a - 2 c - b + c + i \sqrt{3} (b - c)) =$ $A L = \frac{1}{2} (2 a - b - c) + i \frac{\sqrt{3}}{2} (b - c)$ $B M = \frac{1}{2} (2 b - a - c) + i \frac{\sqrt{3}}{2} (c - a)$ $C N = \frac{1}{2} (2 c - a - b) + i \frac{\sqrt{3}}{2} (a - b)$ ${A L}^{2} = \frac{1}{4} {(2 a - b - c)}^{2} + \frac{3}{4} {(b - c)}^{2} =$ $= \frac{1}{4} (4 a^{2} + b^{2} + c^{2} - 4 a b - 4 a c + 2 b c + 3 (b^{2} + c^{2} - 2 b c)) =$ $= \frac{1}{4} (4 a^{2} + 4 b^{2} + 4 c^{2} - 4 a b - 4 a c - 4 b c) =$ $= a^{2} + b^{2} + c^{2} - a b - a c - b c$ $\Rightarrow A L = B M = C N = \sqrt{a^{2} + b^{2} + c^{2} - a b - a c - b c} . ◼$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum