In the triangle $ABC$ given that $\angle ABC = 120^\circ .$ The bisector of $\angle B$ meet $AC$ at $M$ and external bisector of $\angle BCA$ meet $AB$ at $P.$ Segments $MP$ and $BC$ intersects at $K$. Prove that $\angle AKM = \angle KPC .$
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Tags: geometry, angle bisector
FantasyLover
22.01.2011 03:55
Lemma. In $\triangle ABC$, internal angle bisector of $\angle ABC$ and external angle bisectors of $\angle BCA$ and $\angle CAB$ are concurrent.
Proof. It suffices to prove that $B$-excenter $I_B$ lies on the angle bisector of $\angle B$. Let $X$ and $Y$ be the projections of $I_B$ onto $BA$ and $BC$, respectively. In $\triangle BXI_B$ and $\triangle BYI_B$, $BX=BY$ and $I_BX=I_BY$, implying that $\triangle BXI_B\cong \triangle BYI_B$. Hence, $I_B$ lies on the angle bisector of $\angle B$, and the lemma is proven. $\square$
According to above lemma, $P$ is the $M$-excenter of $\triangle MBC$, and therefore $K$ lies on the angle bisector of $\angle BMC$.
By the lemma again, since $K$ is $A$-excenter of $\triangle ABM$, we have that $K$ lies on the angle bisector of $\angle BAC$.
From here on, easy angle chasing shows that $\angle AKM=\angle KPC=30^{\circ}$. We are done. $\blacksquare$
Mahdi_Mashayekhi
17.03.2022 19:40
Note that $BP$ is external bisector of $\angle MBC$ so $P$ is excenter of $BMC$. $\angle CPM = \angle 60 + \angle \frac{A}{2} - \angle CMP = \angle 30$. we have $\frac{CM}{MA} . \frac{AP}{BP} . \frac{BK}{KC} = \frac{CB}{AB} . \frac{CA}{CB} . \frac{BK}{KC} = 1$ so $\frac{BK}{KC} = \frac{AB}{AC}$ so $AK$ is angle bisector of $A$. Now we have $\angle AMK = \angle 30 + \angle \frac{A}{2} - \angle \frac{A}{2} = \angle 30$. We're Done.