Let BD bisect ∠ A B C in Δ A B C. Given ∠ A B C = 60 degrees and A B = B C + C D, find ∠ B A C

Step 1Construct the point E on AB such that E B = B C. Then A E = B A − B E = B A − B C = C D.Besides △ B E D ≅ △ B C D by SAS. Hence E D = C D = A E.Step 2We have ∠ E A D = ∠ E D A = α. ∠ E D B = ∠ C D B = 1 2 ( 180 ∘ − α ).We also have ∠ B E D = 2 α. Now considering the interior angles of △ B E D, 30 ∘ + 2 α + 1 2 ( 180 ∘ − α ) = 180 ∘ .

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Bridger Holden

Answered question

2022-09-29

Let BD bisect

∠ A B C

Δ A B C

. Given

∠ A B C = 60

degrees and

A B = B C + C D

, find

∠ B A C

Answer & Explanation

Alannah Hanson

Beginner2022-09-30Added 11 answers

Step 1

Construct the point E on AB such that

E B = B C

. Then

A E = B A - B E = B A - B C = C D

.
Besides

△ B E D ≅ △ B C D

by SAS. Hence

E D = C D = A E

.
Step 2
We have

∠ E A D = ∠ E D A = α

∠ E D B = ∠ C D B = \frac{1}{2} (180^{\circ} - α)

.
We also have

∠ B E D = 2 α

. Now considering the interior angles of

△ B E D

30^{\circ} + 2 α + \frac{1}{2} (180^{\circ} - α) = 180^{\circ}

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