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Given a triangle ABC, $\angle BAC = 20^{\circ}, \angle ACB=30^{\circ}$. M is a point inside the triangle such that $\angle MAC=\angle MCA=10^{\circ}$. L is a point on AC (L is between A and C) such that $AL=AB$. If $AM \cap BC =K$, prove that $K$ is the center of the excircle of $\triangle ABL$. Find $\angle AMB$.

Proving that $K$ is the excenter of $\triangle ABL$ is easy. However, I cannot find $\angle AMB$.

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  • $\begingroup$ when you write $AM\cap BC=K$, do you mean to say that if you extend the line segment $AM$ to pass through $BC$, it hits point $K$ on $BC$? $\endgroup$
    – Alex R.
    May 7 '12 at 18:55
  • $\begingroup$ I think yes.That's what the OP meant. $\endgroup$ May 7 '12 at 19:04
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    $\begingroup$ It's $140^\circ$, but I used trig (law of sines and observations like that triangle $ABL$ is isosceles, etc.) to find it. $\endgroup$
    – bgins
    May 7 '12 at 19:08
  • $\begingroup$ Yes, $K$ is the intersection point of $AM$ and $BC$. I need a solution without trigonometry. $\endgroup$
    – Adam
    May 7 '12 at 19:11
  • $\begingroup$ I managed to solve it using the law of sines. But I have to solve without it. There must be some way to solve this with additional construction. $\endgroup$
    – Adam
    May 8 '12 at 18:38
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$\hspace{2cm}$enter image description here

Since $\overline{AB}=\overline{AL}$, $\triangle ABL$ is isosceles. $\overline{AK}$ bisects $\angle BAL$; therefore, $\overline{BL}\perp\overline{AK}$. Let $J$ be the intersection of $\overline{BL}$ and $\overline{AK}$. Drop perpendicular $\overline{KN}$ to $\overline{AB}$ and perpendicular $\overline{KP}$ to $\overline{AC}$.

$\angle ABC=130^\circ$ and $\angle ABJ=80^\circ$; therefore, $\angle JBK=50^\circ$. Being an external angle of $\triangle ABC$, $\angle NBK=50^\circ$. Therefore, $\triangle BJK=\triangle BNK$. Thus, $\overline{NK}=\overline{JK}$. $\triangle ANK=\triangle APK$; therefore $$ \overline{PK}=\overline{NK}=\overline{JK} $$ Thus, $K$ is the center of the excircle to $\triangle ABL$ tangent to $\overline{BL}$.


Being an external angle of $\triangle AMC$, $\angle KMC=20^\circ$. Therefore, $\triangle KMC$ is isosceles, giving $\overline{MK}=\overline{KC}$.

Because $\triangle CKP$ is a $30{-}60{-}90$ triangle, we can place $Q$ so that $\triangle CQP$ is also $30{-}60{-}90$ and $\triangle KQC$ is equilateral. Therefore, $\overline{KQ}=\overline{KC}$. Furthermore, $\overline{KP}=\overline{QP}=\frac12\overline{KQ}$. Thus, $$ \frac{\overline{JK}}{\overline{MK}}=\frac{\overline{KP}}{\overline{KQ}}=\frac12 $$ Therefore, $\overline{MJ}=\overline{JK}$, and $\triangle MBK$ is isosceles. Since $\angle MBJ=\angle JBK=50^\circ$, we have that $\angle BMJ=40^\circ$, which leaves $\angle AMB=140^\circ$.

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