Possible length of median and height

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 $$ABC$$.



I don't really see how $$CD$$ helps us.

For (a) do we use the apply the cosine law?

• The first sentence is a bit hard to understand, can you clarify, for example where point $D$ is? Is it a vertex or is it a typo for triangle $ABC$? Feb 25 at 13:34
• I corrected it @dodoturkoz Feb 25 at 13:38
• HINT (for part a): Provided that I drew the diagram correctly... When the height through 𝐶 is drawn a 30-60-90 triangle forms, of which you can find the ratio of the sides easily. For the case where the median is given, you can find $AC'$ via cosine law and use the length of angle bisector theorem (which I couldn't find in english resources): trakademi.com/wp-content/uploads/2020/12/… Feb 25 at 13:55
• With $C'$ I mean the feet of the median on $AB$. Feb 25 at 14:07
• Thank you!! :-) @dodoturkoz Feb 26 at 21:32

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.

• In case i) how did you conclude that $\angle ACD \gt 90^{\circ}$ ? Do you use there the Pythagorean, and since we don't get equality we get this result? Feb 25 at 21:29
• Yes that is correct. For a right angled triangle with two sides as $3$ and $4$, hypotenuse is $5$. As the length of the longest side is more than $5$, the opposite angle will also be more than $90^0$. Feb 26 at 4:40
• I see!! Using law of cosine we get \begin{align*}CM^2=AM^2+AC^2-2\cdot AM\cdot AC\cdot \cos \left (\angle MAC\right ) &\Rightarrow CM^2=\left (\frac{c}{2}\right )^2+b^2-2\cdot \frac{c}{2}\cdot b\cdot \cos \left (30^{\circ}\right )\\ & \Rightarrow CM^2=\left (\frac{2}{2}\right )^2+4^2-2\cdot \frac{2}{2}\cdot 4\cdot \frac{\sqrt{3}}{2} \\ & \Rightarrow CM^2=1+16-4\sqrt{3} \\ & \Rightarrow CM^2=17-4\sqrt{3} \\ & \Rightarrow CM=\sqrt{17-4\sqrt{3}}\approx 3.2\end{align*} Is that correct? As for question (b) : Do we use the formula $R=\frac{BC}{2\sin \left (\angle BAC\right )}$ ? Feb 26 at 20:16
• Yes you calculated $CM$ correctly. Also for (b), yes that is the correct formula to find $R$. Feb 26 at 20:38
• Great!! Thank you very much!! :-) Feb 26 at 21:31