angles in a triangle

Geometry Level 3

In △ A B C \triangle ABC , point D D lies on A C ‾ \overline{AC} such that A B ‾ = A D ‾ \overline{AB}=\overline{AD} . Then ∠ A B C − ∠ A C B = 3 0 ∘ \angle ABC-\angle ACB=30^\circ .Find ∠ C B D \angle CBD in degrees.


The answer is 15.

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1 solution

From the figure, ∠ C B D = ∠ A B C − ∠ A B D \angle CBD=\angle ABC-\angle ABD

Since A D = A B AD=AB , ∠ A B D = ∠ A D B \angle ABD=\angle ADB .

Therefore, we have

∠ C B D = ∠ A B C − ∠ A D B \angle CBD=\angle ABC-\angle ADB

But

∠ A D B = ∠ A C B + ∠ C B D \angle ADB=\angle ACB+\angle CBD

Therefore, we have

∠ C B D = ∠ A B C − ( ∠ A C B + ∠ C B D ) \angle CBD=\angle ABC-(\angle ACB+\angle CBD)

∠ C B D = ∠ A B C − ∠ A C B − ∠ C B D \angle CBD=\angle ABC-\angle ACB-\angle CBD

2 ∠ C B D = ∠ A B C − ∠ A C B 2\angle CBD=\angle ABC-\angle ACB

2 ∠ C B D = 30 2\angle CBD=30

Finally,

∠ C B D = \angle CBD= 1 5 ∘ \large\boxed{15^\circ} answer \color{#69047E}\large\boxed{\text{answer}}

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