Question

$\mathrm{ABC}$ is a triangle in which $\angle \mathrm{B}=2\angle \mathrm{C}$. $\mathrm{D}$ is a point on $\mathrm{BC}$ such that $\mathrm{AD}$ bisects $\angle \mathrm{BAC}$ and $\mathrm{AB}=\mathrm{CD}$. Find $\angle \mathrm{BAC}$?

Answer 1

Step 1. To show that $\mathrm{△}\mathrm{ABP}\cong \mathrm{△}\mathrm{DCP}$.

From $△\mathrm{ABC}$,

Consider $\mathrm{\angle }\mathrm{C}=\mathrm{y}$.

$\mathrm{\angle }\mathrm{B}=2\mathrm{\angle }\mathrm{C}\text{or}\mathrm{\angle }\mathrm{B}=2\mathrm{y}$,

The bisector of $\angle \mathrm{BAC}$ is $\mathrm{AD}$.

Let us consider $\mathrm{\angle }\mathrm{BAD}=\mathrm{\angle }\mathrm{CAD}=\mathrm{x}$.

Assume that $\mathrm{BP}$ is the bisector of $\mathrm{\angle }\mathrm{ABC}$.

From $△\mathrm{BPC}$, we get

$$\begin{gathered} \mathrm{\angle }\mathrm{CBP}=\mathrm{\angle }\mathrm{BCP} \\ \Rightarrow \mathrm{y} \\ \Rightarrow \mathrm{BP}=\mathrm{PC}...\left(\mathrm{i}\right) \end{gathered}$$

so, in $\mathrm{△}\mathrm{ABP}\text{and}\mathrm{△}\mathrm{DCP}$, we have

$$\begin{gathered} \mathrm{\angle }\mathrm{ABP}=\mathrm{\angle }\mathrm{DC} \\ \Rightarrow \mathrm{AB}=\mathrm{DC} \\ \therefore \mathrm{BP}=\mathrm{PC} \end{gathered}$$

As a result, according to the SAS congruence criterion, we obtain

$\mathrm{△}\mathrm{ABP}\cong \mathrm{△}\mathrm{DCP}$.

Step 2. Find $\angle \mathrm{BAC}$

Since,$\mathrm{△}\mathrm{ABP}\cong \mathrm{△}\mathrm{DCP}$

Therefore,

$\mathrm{\angle }\mathrm{BAP}=\mathrm{\angle }\mathrm{CDP}\text{and}\mathrm{AP}=\mathrm{DP}$

$\Rightarrow \mathrm{\angle }\mathrm{ADP}=\mathrm{\angle }\mathrm{DAP}$.

Step 2. Find the value of $\angle \mathrm{BAC}$.

From $△\mathrm{ADB}$, we have

$\begin{array}{r}\mathrm{\angle }\mathrm{ADC}=\mathrm{\angle }\mathrm{ABD}+\mathrm{\angle }\mathrm{BAD}\\ ⟹3\mathrm{x}=2\mathrm{y}+\mathrm{x}\\ \Rightarrow \mathrm{x}=\mathrm{y}\end{array}$

From $△\mathrm{ABC}$, we have

$$\begin{gathered} \mathrm{\angle }\mathrm{A}+\mathrm{\angle }\mathrm{B}+\mathrm{\angle }\mathrm{C}={180}^{\circ } \\ \Rightarrow 2\mathrm{x}+2\mathrm{y}+\mathrm{y}={180}^{\circ } \\ \Rightarrow 5\mathrm{x}={180}^{\circ } \\ \therefore \mathrm{x}={36}^{\circ } \\ \Rightarrow \mathrm{\angle }\mathrm{BAC}=2\mathrm{x} \\ \therefore \mathrm{x}={72}^{\circ } \end{gathered}$$

Therefore, the value of $\angle \mathrm{BAC}$ is $72°$.