Let $D=ML \cap BC$ and $E = KM \cap BC$. Then from conditions on angles $DM=DC$ and $BE=EK$, so $KM +ML = BC \Leftrightarrow DL + ME = DE$. Let $F$ be a point on $BC$ such a $FL \parallel AC$. Then $DL = DF$ and $LF=LB=AM$, so $ALFM$ a parallelogram and $ML \parallel AB$. So $ME=FE$ and $DE=DF+EF=DL+ME$ . $\blacksquare$

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Angle chasing gives us, $\triangle AKM \sim \triangle AML\implies \frac{KM}{ML} = \frac{AM}{AL}\implies\frac{KM+ML}{ML} = \frac{AM+AL}{AL} = \frac{AC}{AL}$. Again, Let $P$ be a point on $AB$ such that $BC=CP$. Then $\angle ACP = 45^{\circ}\implies\triangle ACP\sim\triangle ALM\implies \frac{BC}{ML} = \frac{AC}{AL}$. And we are done.

Let $Z$ be the point such that $MLCZ$ is an isosceles trapezoid. By simple angle chasing we get that $Z$ lies on segment $BC$. Now all that's left to show is $BZ=KM$. But we know $AM=BL$ and $\angle KAM=\angle BLZ$ so we can overly triangles $AKM$ and $BKZ$ with $A$ and $L$ matching up, $M$ and $B$ matching up, and $K$ lying on segment $LZ$. It is not hard to show that $BZK$ is isosceles implying the result.

From point I, we draw a line parallel to the straight line AC, intersecting the side BC at the point N. As a result, NLMC is an isosceles trapezoid, which means that LM=MC. A parallelogram NLAM is formed from AM=BL=LN, LN||AM. From this it follows that BK=NM and BKNM is an equilateral trapezoid. The statement to be proved from BN=KM and LM=MC yields KM+ML=BC.

Let $T$ be a point on $KM$ with $ML = MT$. $\angle MLT =\frac{ \angle KML}{2} = \angle 30$ which gives $\angle 75 = \angle ALT = \angle LKM = \angle ABC$ implying $BC \parallel LT$ and $KTL$ is an isosceles triangle. By using $AKM \sim AML$, $\frac{MT}{MK} = \frac{ML}{MK} = \frac{AM}{AL} = \frac{CM}{AM}$ and thus $CT \parallel AK$. At the end, $BLTC$ turns out to be a parallelogram. This implies $BC = LT = KT = KM + MT = KM + ML$ as desired. $\square$

NO_SQUARES wrote: Let $D=ML \cap BC$ and $E = KM \cap BC$. Then from conditions on angles $DM=DC$ and $BE=EK$, so $KM +ML = BC \Leftrightarrow DL + ME = DE$. Let $F$ be a point on $BC$ such a $FL \parallel AC$. Then $DL = DF$ and $LF=LB=AM$, so $ALFM$ a parallelogram and $ML \parallel AB$. So $ME=FE$ and $DE=DF+EF=DL+ME$ . $\blacksquare$ In which app you guys draw these geometrical figures?

Rohit-2006 wrote: In which app you guys draw these geometrical figures? Looks a bit like Geogebra but I could be wrong