## Monday, January 19, 2015

### Geometry Problem 1075: Quadrilateral, Right Triangle, 90 Degrees, Isosceles, Midpoint, Distance

Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

1. http://s22.postimg.org/sj3p1a4o1/pro_1075.png

Draw 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= 6
In triangle BDL, M and N are midpoints of BD and BL= > MN= 1/2 DL=3

2. 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

Let 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 = <MDN
Thus by SAS ΔDMN is congruent to ΔNLD and
hence MN = LD = ½ AD = 3

4. 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 = 3

Sumith Peiris
Moratuwa
Sri Lanka