Ram drew $∠ A B C$ and then he drew a line segment $B D$ such that line segment $B D$ is angle bisector of $∠ A B C$ . Then, $∠ A B C$ $=$ ___ $∠ A B D$ .

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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, $∠ A B D + ∠ A B D = ∠ A B C$ (As, $∠ A B D = ∠ D B C$ )

$\Rightarrow 2 ∠ A B D = ∠ A B C$
Hence, $∠ A B C = 2 ∠ A B D$

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Ram drew ∠ABC and then he drew a line segment BD such that 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 = ∠ABCSo, ∠ABD+∠ABD=∠ABC (As, ∠ABD=∠DBC) ⇒2∠ABD=∠ABCHence, ∠ABC=2∠ABD

Ram drew $∠ A B C$ and then he drew a line segment $B D$ such that line segment $B D$ is angle bisector of $∠ A B C$ . Then, $∠ A B C$ $=$ ___ $∠ A B D$ .

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, $∠ A B D + ∠ A B D = ∠ A B C$ (As, $∠ A B D = ∠ D B C$ )

$\Rightarrow 2 ∠ A B D = ∠ A B C$
Hence, $∠ A B C = 2 ∠ A B D$