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Question-71149




Question Number 71149 by naka3546 last updated on 12/Oct/19
Commented by MJS last updated on 17/Oct/19
if the hexagon is regular it′s symmetric  BE is the symmetry axis  M∈(BE)  A=C′ ∧ F=D′  ⇒ (AF)=(CD)′ and similar for all pairs of lines
$$\mathrm{if}\:\mathrm{the}\:\mathrm{hexagon}\:\mathrm{is}\:\mathrm{regular}\:\mathrm{it}'\mathrm{s}\:\mathrm{symmetric} \\ $$$${BE}\:\mathrm{is}\:\mathrm{the}\:\mathrm{symmetry}\:\mathrm{axis} \\ $$$${M}\in\left({BE}\right) \\ $$$${A}={C}'\:\wedge\:{F}={D}' \\ $$$$\Rightarrow\:\left({AF}\right)=\left({CD}\right)'\:\mathrm{and}\:\mathrm{similar}\:\mathrm{for}\:\mathrm{all}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{lines} \\ $$
Commented by naka3546 last updated on 12/Oct/19
Shaded  area  is  ...
$${Shaded}\:\:{area}\:\:{is}\:\:… \\ $$
Commented by MJS last updated on 13/Oct/19
M∈BE ⇒ we get 3 pairs of symmetric triangled  ⇒ areas MAB, MCD, MEF are the same as  areas MBC, MAF, MDE
$${M}\in{BE}\:\Rightarrow\:\mathrm{we}\:\mathrm{get}\:\mathrm{3}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{symmetric}\:\mathrm{triangled} \\ $$$$\Rightarrow\:\mathrm{areas}\:{MAB},\:{MCD},\:{MEF}\:\mathrm{are}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as} \\ $$$$\mathrm{areas}\:{MBC},\:{MAF},\:{MDE} \\ $$
Commented by naka3546 last updated on 14/Oct/19
how  to  prove  them  are  symmetric  triangled  ?
$${how}\:\:{to}\:\:{prove}\:\:{them}\:\:{are}\:\:{symmetric}\:\:{triangled}\:\:? \\ $$
Answered by MJS last updated on 12/Oct/19
shaded area is half of the hexagon
$$\mathrm{shaded}\:\mathrm{area}\:\mathrm{is}\:\mathrm{half}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hexagon} \\ $$
Commented by naka3546 last updated on 13/Oct/19
could  it  proved  if  shaded  area  is  a half  hexagone?
$${could}\:\:{it}\:\:{proved}\:\:{if}\:\:{shaded}\:\:{area}\:\:{is}\:\:{a}\:{half}\:\:{hexagone}? \\ $$

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