Geometry Problem

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

Click the figure below to see the complete problem 762.

## Friday, June 8, 2012

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

Labels:
congruence,
midpoint,
parallelogram,
perpendicular

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MC cut AD at N

ReplyDeleteTriangle MBc congruence to triangle MAN ... ( case ASA)

so NA=BC=AD

in right triangle NHD, HA is a median => AH=AN=AD

Extend DA and CM to meet at E.

ReplyDeleteTriangles 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)

Using vector approach to solve the problem. Let vector DA = a, vector DC = c.

ReplyDeleteWe 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.

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

ReplyDeleteu.u

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

ReplyDeleteLet N be mid point of CD.

ReplyDeleteMCNA 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

A variant on above:

ReplyDeleteLet M' be the mid point of CD. AM' is || to MC and E is the intersection of DH with AM'.

1. angle DEM' is 90 degrees like DHC since its a traversal through the 2 parallel lines.

2. Therefore tr. DEM' is similar to tr. DHC (AAA) in a 2x ratio. So EH = DE.

3. tr AHE is congruent to tr AED by SAS therefore AH = AD.

N be the mid-point of CD.

ReplyDeleteHN=CN=DN=AM and AN//MH => AMHN is isosceles trapezoid

=>MN=AH=AD