## Thursday, September 22, 2016

### Geometry Problem 1262: Isosceles Triangle, Altitude, Medians, 90 Degrees, Perpendicular

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Omid Motahed.

Click on the figure below to view more details of problem 1262. 1. Let the centroid be G.
Since BG:GH=2:1, thus AH=GH=CH.
Therefore H is the Orthocenter if AGC, with diameter AC.
Consequently, ∠AGC=90°

1. Typo: H is the circumcenter of AGC

2. Palai Renato

BH/AC=1,5 we put AC=2 BH=3, BC= √10, CD= √10/2 , CD2= 2,5
The cross EC with AD =K; CK=X ; X2+X2=4, X=√2
ED=1 , KD=y ; y2+y2=1 y=1/√2
CD2= 2+1/2 = 2,5 therefore EC and AD are perpendicular

3. Let AD, CE meet at M and let AC = 2p

Then BH = 3p and MH = p
So AH = MH = CH = p

Hence < AMH = < CMH = 45 and so AD is perpendicular to CE

Sumith Peiris
Moratuwa
Sri Lanka

4. Problem 1262
BH is altitude and median then K is centroid of the triangle ABC with BH=3.KH. Then
3.KH=3.AC/2 or KH=AC/2 (AH=HC).Therefore AD is perpendicular CE.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

5. Comment sent by Sumith Peiris:

Dear Omid and Antonio

Just realized...,

The triangle need not be isoceles for the medians AD and CE to be perpendicular.

It is sufficient that the median BH (as opposed to altitude) be 1.5 times AC

Hope u will adjust the problem accordingly.

Rgds

Sumith

1. Thanks Sumith
See problem 1263 at:
http://www.gogeometry.com/school-college/3/p1263-triangle-medians-90-degree-perpendicular.htm

6. @Sumith, thanks that makes the problem even more interesting

7. extend AC to AZ such that HB=HZ
then LBZH=45.......so....LECH=45 etc.