Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
http://s22.postimg.org/sj3p1a4o1/pro_1075.pngDraw circle centered N radius NB and diameter BL of circle N ( see sketch)Note that ND is perpendicular bisector of BC and angle (AND)=angle (DNL)Triangles AND congruence to LDN ( case SAS) so DL=AD= 6In triangle BDL, M and N are midpoints of BD and BL= > MN= 1/2 DL=3
Fie PC perpendicular pe BC,P fiind de aceeasi parte a dreptei BC cu A, astfel incat PC=AB rezulta ca ABCP dreptunghi,AD=DC=> N mijlocul diagonalei BPa dreptunghiului=>MN linie mijlocie a triunghiului BPD =>MN=PD/2=6/2=3
http://www.uploaddata.net/uploaddata/pravin5880/GoGeometry%20Problem%201075.pngLet K, L, M, N be the midpoints of BC, AD, BD, AC respectively.LN // DC // MK, LN = ½ DC = ½ DB = DM. Also LN = ½ DC= MK.Follows LNKM is a parallelogram and <DNL = <NKM = <DKM = <MDK = <MDNThus by SAS ΔDMN is congruent to ΔNLD and hence MN = LD = ½ AD = 3
Let MN meet DC at P. NB = NC so MD = PD = PC and MN = PN. Now MP is // to AD by applying mid point theorem to Tr. ADC Hence MN = NP = AD/2 = 3Sumith PeirisMoratuwaSri Lanka