Monday, January 19, 2015

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

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.

Online Math: Geometry Problem 1075: Quadrilateral, Right Triangle, 90 Degrees, Isosceles, Midpoint, Distance.

Posted by Antonio Gutierrez at 9:07 PM
Labels: , , , ,

4 comments:

  1. Peter TranJanuary 20, 2015 at 12:14 PM

    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

    ReplyDelete
  2. ion raduJanuary 21, 2015 at 11:32 AM

    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

    ReplyDelete
  3. PravinJanuary 25, 2015 at 7:09 PM

    http://www.uploaddata.net/uploaddata/pravin5880/GoGeometry%20Problem%201075.png

    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

    ReplyDelete
  4. Sumith PeirisJanuary 18, 2016 at 7:59 AM

    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

    ReplyDelete

Subscribe to: Post Comments (Atom)