Let BD bisect $\mathrm{\angle}ABC$ in $\mathrm{\Delta}ABC$. Given $\mathrm{\angle}ABC=60$ degrees and $AB=BC+CD$, find $\mathrm{\angle}BAC$

Bridger Holden
2022-09-29
Answered

Let BD bisect $\mathrm{\angle}ABC$ in $\mathrm{\Delta}ABC$. Given $\mathrm{\angle}ABC=60$ degrees and $AB=BC+CD$, find $\mathrm{\angle}BAC$

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Alannah Hanson

Answered 2022-09-30
Author has **11** answers

Step 1

Construct the point E on AB such that $EB=BC$. Then $AE=BA-BE=BA-BC=CD$.

Besides $\mathrm{\u25b3}BED\cong \mathrm{\u25b3}BCD$ by SAS. Hence $ED=CD=AE$.

Step 2

We have $\mathrm{\angle}EAD=\mathrm{\angle}EDA=\alpha $. $\mathrm{\angle}EDB=\mathrm{\angle}CDB=\frac{1}{2}({180}^{\circ}-\alpha )$.

We also have $\mathrm{\angle}BED=2\alpha $. Now considering the interior angles of $\mathrm{\u25b3}BED$, ${30}^{\circ}+2\alpha +\frac{1}{2}({180}^{\circ}-\alpha )={180}^{\circ}$.

Construct the point E on AB such that $EB=BC$. Then $AE=BA-BE=BA-BC=CD$.

Besides $\mathrm{\u25b3}BED\cong \mathrm{\u25b3}BCD$ by SAS. Hence $ED=CD=AE$.

Step 2

We have $\mathrm{\angle}EAD=\mathrm{\angle}EDA=\alpha $. $\mathrm{\angle}EDB=\mathrm{\angle}CDB=\frac{1}{2}({180}^{\circ}-\alpha )$.

We also have $\mathrm{\angle}BED=2\alpha $. Now considering the interior angles of $\mathrm{\u25b3}BED$, ${30}^{\circ}+2\alpha +\frac{1}{2}({180}^{\circ}-\alpha )={180}^{\circ}$.

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