## 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.

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

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)

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.

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

u.u

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

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

7. A variant on above:

Let 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.

8. N be the mid-point of CD.
HN=CN=DN=AM and AN//MH => AMHN is isosceles trapezoid