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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?

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    $\begingroup$ 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 $ABC$? $\endgroup$
    – dodoturkoz
    Feb 25 at 13:34
  • $\begingroup$ I corrected it @dodoturkoz $\endgroup$
    – Mary Star
    Feb 25 at 13:38
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    $\begingroup$ 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/… $\endgroup$
    – dodoturkoz
    Feb 25 at 13:55
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    $\begingroup$ With $C'$ I mean the feet of the median on $AB$. $\endgroup$
    – dodoturkoz
    Feb 25 at 14:07
  • $\begingroup$ Thank you!! :-) @dodoturkoz $\endgroup$
    – Mary Star
    Feb 26 at 21:32
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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 $\angle B \gt 90^0$.

enter image description here

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.

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  • $\begingroup$ 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? $\endgroup$
    – Mary Star
    Feb 25 at 21:29
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    $\begingroup$ 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 $90^0$. $\endgroup$
    – Math Lover
    Feb 26 at 4:40
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    $\begingroup$ 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 $R=\frac{BC}{2\sin \left (\angle BAC\right )}$ ? $\endgroup$
    – Mary Star
    Feb 26 at 20:16
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    $\begingroup$ Yes you calculated $CM$ correctly. Also for (b), yes that is the correct formula to find $R$. $\endgroup$
    – Math Lover
    Feb 26 at 20:38
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    $\begingroup$ Great!! Thank you very much!! :-) $\endgroup$
    – Mary Star
    Feb 26 at 21:31

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