In the figure, $∠ A B C = 80^{\circ}, ∠ A B C$ is bisected and the resultant angles are bisected again.

Then, $∠ A B D + ∠ A B F - (∠ E B F + ∠ D B C + ∠ F B C)$ is equal to:

60^{\circ}

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$20^{\circ}$

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$40^{\circ}$

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$0^{\circ}$

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Solution

The correct option is D
$0^{\circ}$

$∠ A B D = 40^{\circ}$

$∠ A B F = 60^{\circ}$

$∠ E B F = 40^{\circ}$

$∠ D B C = 40^{\circ}$

$∠ F B C = 20^{\circ}$

So, $∠ A B D + ∠ A B F - (∠ E B F + ∠ D B C + ∠ F B C) = 40^{\circ} + 60^{\circ} - (40^{\circ} + 40^{\circ} + 20^{\circ}) = 100^{\circ} - 100^{\circ} = 0^{\circ}$

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Question 40

In $Δ A B C$ , $∠ A = 100^{\circ}$ , AD bisects $∠ A$ and $A D ⊥ B C$ . Then, $∠ B$ is equal to

a) $80^{\circ}$
b) $20^{\circ}$
c) $40^{\circ}$
d) $30^{\circ}$

In the given figure, $∠ A = 60^{\circ}$ and $∠ A B C = 80^{\circ}$ , then $∠ D P C$ and $∠ B Q C$ are respectively ___.

In ΔABC, BC = AB and ∠B = 80°, then ∠A is equal to:

Bisect the given angles and measure the bisected angles in the following:

∠ABC = 60°

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