In a triangle $ABC$ let $D$ be the intersection point of the side $AB$ with the angle bisector of the inner angle at $C$.
It holds that $|AC|=4$, $|CD|=3$.
Let the inner angle at $A$ be equal to $\frac{\pi}{6}$.
(a) Which length are possible for the median through $C$ and the height through $C$ of the triangle $ABC$ ?
(b) In each case calculate the legth of the radius of the circumscribed circle of $ABC$.
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I don't really see how $CD$ helps us.
For (a) do we use the apply the cosine law?
If $CH$ is the height of $\triangle ABC$ through $C$, $AB = c, BC = a, AC = b$,
$CH = b \displaystyle \sin 30^0 = 2$
Applying Pythagoras,
$AH = \sqrt{AC^2 - CH^2} = 2 \sqrt3$
$DH = \sqrt{CD^2 - CH^2} = \sqrt5$
i) If $H$ is between $A$ and $D, AD = 2\sqrt3 + \sqrt5 \gt 5$.
In $\triangle ACD, AC = 4, CD = 3, AD \gt 5$, which means $\angle ACD \gt 90^0$ but $CD$ is internal angle bisector hence $\angle ACD$ must be less than $90^0$. So we conclude $H$ is not in between $A$ and $D$.
ii) If $D$ is between $A$ and $H, AD = AH - DH = 2 \sqrt3 - \sqrt5 \lt 3$.
In $\triangle ACD, AD$ is the smallest side and hence $\angle ACD \lt 30^0$ which means $\angle C \lt 60^0$ and hence $\triangle ABC$ is obtuse angled triangle with $\angle B \gt 90^0$.
If $BH = x$, applying angle bisector theorem,
$\displaystyle \frac{4}{a} = \frac{2 \sqrt3 - \sqrt5}{\sqrt5 - x}$ ...($1$)
Applying Pythagoras, we also have $x^2 + 2^2 = a^2$ ...($2$)
Solving $(1)$ and $(2)$ and I leave further workings to you, we get $a \approx 2.5, b = 4$ and $c \approx 2$. Knowing the sides, you can apply the formula for median in terms of the side lengths of $\triangle ABC$ or you can apply law of cosine.