Exam

Olympiad workbook

Subject

IOQM

Chapter

Properties of Triangles

Olympiad workbook · IOQM · Properties of Triangles

In a triangle A B C , the median A D divides B A C in the ratio 1:...

In a triangle $A B C$, the median $A D$ divides $\angle B A C$ in the ratio $1: 2$. Extend $A D$ to $E$ such that $E B$ is perpendicular $A B$. Given that $B E=3, B A=4$, find the integer nearest to $B C^2$.

Step-by-step solution

Here, $D$ is mid-point of $B C$, hence $B D: C D=1: 1$ $\angle A B E=90^{\circ}$ Let $\angle B A D=\theta$, then $\angle C A D=2 \theta$ $\therefore \quad \tan \theta=\frac{3}{4}, \text { and } \tan 2 \theta=\frac{24}{7}$ Now, using $\cot m-n$ theorem in $\triangle A B C$ $$\begin{aligned} 2 \cot \alpha & =\cot \theta-\cot 2 \theta \\ \Rightarrow \quad \cot \alpha & =\frac{25}{48} \end{aligned}$$ Now, using sine rule in $\triangle A B D$, we get $$\begin{aligned} & \frac{B D}{4}=\frac{\sin \theta}{\sin (\pi-\alpha)} \\ \therefore & B D=\frac{4 \times 3 \sqrt{25^2+48^2}}{5 \times 48} \end{aligned}$$ So, $4 B D^2=B C^2=\frac{25^2+48^2}{100}=29.29$ Nearest integer is 29 .