Ram drew $∠ A B C$ and then he drew a line segment $B D$ so that the line segment $B D$ is angle bisector of $∠ A B C$ . Then, $∠ (A B C) =$ __ $\times ∠ (A B D)$ .

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Solution

As stated in question line segment $B D$ is angle bisector so, $∠ A B D = ∠ D B C$

And, $∠ A B D + ∠ D B C = ∠ A B C$

So, $∠ (A B C) = 2 ∠ (A B D)$

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Ram drew $∠ A B C$ and then he drew a line segment $B D$ so that the line segment $B D$ is angle bisector of $∠ A B C$ . Then, $∠ (A B C) =$ __ $\times ∠ (A B D)$ .

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Ram drew ∠ABC and then he drew a line segment BD so that the line segment BD is angle bisector of ∠ABC. Then, ∠(ABC)= __ ×∠(ABD).

As stated in question line segment BD is angle bisector so, ∠ABD=∠DBC And, ∠ABD+∠DBC=∠ABC So,∠(ABC)=2∠(ABD)

Ram drew $∠ A B C$ and then he drew a line segment $B D$ so that the line segment $B D$ is angle bisector of $∠ A B C$ . Then, $∠ (A B C) =$ __ $\times ∠ (A B D)$ .

As stated in question line segment $B D$ is angle bisector so, $∠ A B D = ∠ D B C$

And, $∠ A B D + ∠ D B C = ∠ A B C$

So, $∠ (A B C) = 2 ∠ (A B D)$