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Answer 1 · 2017-07-31T04:12:46

$∠ A C B = 100^{\circ}$ $angleACB=100^@$

Explanation:

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Let $∠ D E F = x, and ∠ D F E = y$ $angleDEF=x, and angleDFE=y$
$\Rightarrow x + y = 180 - 40 = 140^{\circ}$ $=> x+y=180-40=140^@$

Given $E C = A C, \Rightarrow ∠ E A C = x$ $EC=AC, => angleEAC=x$
$\Rightarrow ∠ E C A = 180 - 2 x$ $=> angleECA=180-2x$
Given $C F = B C, \Rightarrow ∠ F B C = y$ $CF=BC, => angleFBC=y$
$\Rightarrow ∠ F C B = 180 - 2 y$ $=> angleFCB=180-2y$

$∠ A C B = 180 - (180 - 2 x + 180 - 2 y)$ $angleACB=180-(180-2x+180-2y)$
$= 2 (x + y) - 180$ $=2(x+y)-180$
$= 2 \cdot (140) - 180 = 280 - 180 = 100^{\circ}$ $=2*(140)-180=280-180=100^@$

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