The perpendicular $AD$ to the base of a $\triangle ABC$ divides it into segments such that $BD$, $CD$ and $AD$ are in the ratio of $2$, $3$ and $6$. Prove that the vertical angle of the triangle is $45^\circ$.
Let $ABC$ be the given triangle with base $BC$.
$AD$ is the perpendicular to the base of the triangle.
Given: $\;$ $BD : CD : AD = 2 : 3 : 6$
$\therefore \;$ Let $\;$ $BD = 2 k$, $\;$ $CD = 3 k$, $\;$ $AD = 6 k$ $\;$ where $k$ is the constant of proportionality
To prove that: $\;$ In $\triangle ABC$, $\;$ $\angle A = 45^\circ$
In $\triangle ACD$,
$\begin{aligned}
AC & = \sqrt{CD^2 + AD^2} \\\\
& = \sqrt{\left(3 k\right)^2 + \left(6 k\right)^2} \\\\
& = \sqrt{9 k^2 + 36 k^2} \\\\
& = 3 k \sqrt{5}
\end{aligned}$
Applying sine rule to $\triangle ACD$, we have
$\dfrac{CD}{\sin \left(\angle DAC\right)} = \dfrac{CA}{\sin \left(\angle CDA\right)}$
i.e. $\;$ $\dfrac{3 k}{\sin \left(\angle DAC\right)} = \dfrac{3 k \sqrt{5}}{\sin 90^{\circ}}$
i.e. $\;$ $\sin \left(\angle DAC\right) = \dfrac{1}{\sqrt{5}}$
$\implies$ $\angle DAC = \sin^{-1} \left(\dfrac{1}{\sqrt{5}}\right)$ $\;\;\; \cdots \; (1)$
In $\triangle ADB$,
$\begin{aligned}
AB & = \sqrt{DB^2 + AD^2} \\\\
& = \sqrt{\left(2 k\right)^2 + \left(6 k\right)^2} \\\\
& = \sqrt{4 k^2 + 36 k^2} \\\\
& = 2 k \sqrt{10}
\end{aligned}$
Applying sine rule to $\triangle ADB$, we have
$\dfrac{DB}{\sin \left(\angle DAB\right)} = \dfrac{AB}{\sin \left(\angle ADB\right)}$
i.e. $\;$ $\dfrac{2 k}{\sin \left(\angle DAB\right)} = \dfrac{2 k \sqrt{10}}{\sin 90^{\circ}}$
i.e. $\;$ $\sin \left(\angle DAB\right) = \dfrac{1}{\sqrt{10}}$
$\implies$ $\angle DAB = \sin^{-1} \left(\dfrac{1}{\sqrt{10}}\right)$ $\;\;\; \cdots \; (2)$
Now, in $\triangle ABC$,
$\begin{aligned}
\angle A & = \angle DAC + \angle DAB \\\\
& = \sin^{-1} \left(\dfrac{1}{\sqrt{5}}\right) + \sin^{-1} \left(\dfrac{1}{\sqrt{10}}\right) \;\; \left[\text{from equations (1) and (2)}\right] \\\\
& \left[\text{Note: } \sin^{-1} x + \sin^{-1} y = \sin^{-1} \left(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\right)\right] \\\\
& = \sin^{-1} \left(\dfrac{1}{\sqrt{5}} \times \sqrt{1 - \dfrac{1}{10}} + \dfrac{1}{\sqrt{10}} \times \sqrt{1 - \dfrac{1}{5}}\right) \\\\
& = \sin^{-1} \left(\dfrac{1}{\sqrt{5}} \times \dfrac{3}{\sqrt{10}} + \dfrac{1}{\sqrt{10}} \times \dfrac{2}{\sqrt{5}}\right) \\\\
& = \sin^{-1} \left(\dfrac{5}{5 \sqrt{2}}\right) \\\\
& = \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right)
\end{aligned}$
$\implies$ $\angle A = 45^{\circ}$
Hence proved.