The median $BM$ is drawn in the triangle $ABC$. It is known that $\angle ABM = 40 {} ^ \circ$ and $\angle CBM = 70 {} ^ \circ $ Find the ratio $AB: BM$.
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Tags: ratio, angles, median, geometry
sunken rock
18.08.2020 10:43
$BC$ is external angle bisector of $\angle ABM$, done. Best regards, sunken rock
yofro
18.08.2020 20:47
sunken rock wrote: $BC$ is external angle bisector of $\angle ABM$, done. Best regards, sunken rock Could you elaborate?
aa1024
18.08.2020 21:28
Denote, $\angle{BAM} = \gamma$ and $\angle{BMC} = \alpha$
By the Law Of Sines in $\triangle{MCB}$ we obtain,
$$\dfrac{\sin {70}}{MC} = \dfrac{\sin{70 - \gamma}}{BM} = \dfrac{\sin {\alpha}}{BC}.$$We also have, by LoS on $\triangle{ABM}$ ,
$$\dfrac{AC}{\sin{40}} = \dfrac{BM}{\sin{\gamma}} = \dfrac{AB}{\sin{\alpha}}.$$Since $BM$ is a median we have $AM = MC.$ Thus $$\dfrac{\sin{70}}{\sin{40}} = \dfrac{AB}{BC} = \dfrac{\sin{70 - \gamma}}{\sin{\gamma}}.$$So WLOG let $$AB = \sin{70}$$and $$BC = \sin{40}.$$However we can find $AC$ by Law Of Cosines in $\triangle{ABC}.$ (Because of our WLOG statement). This turn out to be $$AC^2 = \sin^2{70} + \sin^2{40} - 2\sin{70}\sin{40}\cos{110}.$$So $$AC = \sqrt{\sin^2{70} + \sin^2{40} - 2\sin{70}\sin{40}\cos{110}}.$$We can divide by $2$ and now use Stewarts formula in reverse to find the length of median $BM$ which will give the answer. However this method is very bashy.
rafaello
18.08.2020 22:31
yofro wrote: Could you elaborate? Proof of exterior angle bisector theorem, basically: Extend the diagram such that $E$ lies on ray $AB$ and $BE=BM$. Thus, by symmetry $\angle BCE=\angle BCM$, hence we can use interior angle bisector theorem and we get that $$\frac {AC} {CE}=\frac {AB} {BE},$$which is same as $$\frac {AC} {CM}=\frac {AB} {BM},$$since $CM=CE$ and $BM=BE$ by symmetry. We know that $\frac {AC} {CM}=2$, hence $$\frac {AB} {BM}=2.$$
aa1024
18.08.2020 22:32
rafaello wrote: yofro wrote: Could you elaborate? Proof of exterior angle bisector theorem, basically: Extend the diagram such that $E$ lies on ray $AB$ and $BE=BM$. Thus, by symmetry $\angle BCE=\angle BCM$, hence we can use interior angle bisector theorem and we get that $$\frac {AC} {CE}=\frac {AB} {BE},$$which is same as $$\frac {AC} {CM}=\frac {AB} {BM},$$since $CM=CE$ and $BM=BE$ by symmetry. We know that $\frac {AC} {CM}=2$, hence $$\frac {AB} {BM}=2.$$ That is a much cleaner solution than the one I had!