DA=DK, AM=NA Let DAK = x => DAN= 180 - x, DKA = x, ADB = x, => DBA = 180 - 2x since DKCB is isosceles trapezoid CKB= 180 - 2x => DKM = 180 - x. so ΔNAD ≅ ΔMKD

I don't understand the solution. Could someone please explain it to me in detail ?? I request the others to please verify if this will be the diagram for this question - Diagram - [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.516002582362034, xmax = 7.137730306759368, ymin = -3.6170251934599027, ymax = 6.384996468288456; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-1.89,1.47)--(-3.93,-2.37)--(-2.51,-3.39)--(-0.43,0.45)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-1.89,1.47)--(-3.93,-2.37), linewidth(2) + rvwvcq); draw((-3.93,-2.37)--(-2.51,-3.39), linewidth(2) + rvwvcq); draw((-2.51,-3.39)--(-0.43,0.45), linewidth(2) + rvwvcq); draw((-0.43,0.45)--(-1.89,1.47), linewidth(2) + rvwvcq); draw((-3.93,-2.37)--(-0.43,0.45), linewidth(2) + wrwrwr); draw((-2.221349686664511,0.846282942749155)--(-0.43,0.45), linewidth(2) + wrwrwr); draw((-0.43,0.45)--(-2.0993683510452095,4.691809772741752), linewidth(2) + wrwrwr); draw((-0.43,0.45)--(0.4860190793863558,5.8714452338221585), linewidth(2) + wrwrwr); /* dots and labels */ dot((-1.89,1.47),dotstyle); label("$A$", (-1.7882792446786815,1.7141257167140391), NE * labelscalefactor); dot((-3.93,-2.37),dotstyle); label("$B$", (-3.8424841825438762,-2.125281131438754), NE * labelscalefactor); dot((-2.51,-3.39),dotstyle); label("$C$", (-2.424104582589337,-3.152383600371348), NE * labelscalefactor); dot((-0.43,0.45),dotstyle); label("$D$", (-0.3209900033463994,0.6870232477814449), NE * labelscalefactor); dot((-2.221349686664511,0.846282942749155),dotstyle); label("$K$", (-2.1306467343228808,1.1027551994922569), NE * labelscalefactor); dot((-2.0993683510452095,4.691809772741752),dotstyle); label("$M$", (-2.0083726308785237,4.942162047645049), NE * labelscalefactor); dot((0.4860190793863558,5.8714452338221585),dotstyle); label("$N$", (0.5838383621418413,5.993719337266515), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

$\triangle DAK\sim \triangle DBC$. So by Gliding principle $\triangle DNM$ is also similar to them. Hence the result. @above: It seems that you forgot condition $AB=BD$. @below: Your diagram. On it clearly $DM$ and $DN$ are not equal. I thought that was your concern.

zuss77 wrote: $\triangle DAK\sim \triangle DBC$. So by Gliding principle $\triangle DNM$ is also similar to them. Hence the result. @above: It seems that you forgot condition $AB=BD$. Thanks @above I got what he was trying to say ... But I didn't forget about the condition $AB=BD$. What made you feel that ??

Here is nice solution. $\angle BKD=180^{\circ}-\angle DKA=180^{\circ}-\angle DAB=\angle ABC$, and $DKBC$ is isosceles trapezoid, so $AD=DK=BC$ $\implies$ triangle $DKB$ is congruent with triangle $KBC$ $\implies CK=DB=AB=CD$, $\angle NBD=\angle KDB=\angle BKC=\angle DCM$ Now we have $CM=2CK=2DB=2AB=NB \implies \angle DCM=\angle NBD, DC=DB \implies$ triangle $CDM$ is congruent with triangle $BDM$ $\implies$ $DM=DN$. $Q.E.D.$