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Let lines $A B$ and $C D$ intersect at $O .$ If $∠ A O C + ∠ B O E = 70^{\circ}$ and $∠ B O D = 40^{\circ},$ find $∠ B O E$ .

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Solution

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$∠ A O G + ∠ B O E = 70^{\circ}$
$∠ D O B = ∠ A O C = 40^{\circ}$
$∴ 40^{\circ} + ∠ B O E = 70^{\circ}$
$∠ B O C = 70 - 40$
$= 30^{\circ}$

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