The base of an isosceles triangle is $a$, and the angle at the vertex is $\alpha$. Find the bisector drawn to the lateral side

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Apply the Law of Sines.

## Solution

Let $CD$ be the bisector of isosceles triangle $ABC (AB = AC)$, $\angle A = \alpha, BC = a$. In triangle $CBD$, $\angle DBC = 90^\circ - \alpha / 2$,

$\angle DCB = 1/2 \left(90^\circ - \alpha / 2\right) = 45^\circ - \alpha / 4, \angle BDC = 180^\circ - \left(90^\circ - \alpha / 2\right) - \left(45^\circ - \alpha / 4\right) = 45^\circ + 3\alpha / 4, BC = a$.

By the Law of Sines, $CD = \frac{BC \sin \angle DBC}{\sin \angle BDC} = \frac{a \cos \frac{\alpha}{2}}{\sin \left(45^\circ + \frac{3\alpha}{4}\right)}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5ddd4eab4fe1a12edeb9g-13.jpg?height=420&width=486&top_left_y=1711&top_left_x=795)

Answer

$\frac{a \cos \frac{\alpha}{2}}{\sin \left(45^\circ + \frac{3\alpha}{4}\right)} \cdot$