Friday, June 8, 2012

Problem 762: Parallelogram, Midpoint, Perpendicular, Congruence, Isosceles triangle

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

Click the figure below to see the complete problem 762.

Online Geometry Problem 762: Parallelogram, Midpoint, Perpendicular, Congruence, Isosceles triangle.

6 comments:

  1. MC cut AD at N
    Triangle MBc congruence to triangle MAN ... ( case ASA)
    so NA=BC=AD
    in right triangle NHD, HA is a median => AH=AN=AD

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  2. Extend DA and CM to meet at E.
    Triangles BMC and AEM are similar
    So BC/AE = BM/MD = 1
    Follows AE = BC = AD
    Triangle HED is right angled at H
    and A is the midpoint of hypotenuse ED
    Hence AH = AD (= AE)

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  3. Using vector approach to solve the problem. Let vector DA = a, vector DC = c.
    We have vector MC = 1/2 c - a
    Let vector DH = h, such that DH。MC = 0, => 1/2(h。c)= h。a
    Now |AH|。|AH|
    =(h-a)。(h-a)
    = |h|。|h| + |a|。|a| - 2(h。a)
    = |h|。|h| + |a|。|a| - (h。c)
    = |a|。|a| + |h|(|h|-|c|sinX), where X is the angle between DH and DC
    = |a|。|a|
    That means |AH| = |a| = |AD|
    Q.E.D.

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  4. Challenging problems in geometry (chaper 1, problem 16)...

    u.u

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  5. http://www.mathematica.gr/forum/viewtopic.php?f=20&t=27390&p=134010

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  6. Let N be mid point of CD.

    MCNA is a parellllogram and AMHN is an isoceles trapezoid whose diagonals are equal MN = AH. But MN = AD since ADNM is a parallelogram and so AH = AD

    Sumith Peiris
    Moratuwa
    Sri Lanka

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