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.

## Thursday, September 22, 2016

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

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Let the centroid be G.

ReplyDeleteSince BG:GH=2:1, thus AH=GH=CH.

Therefore H is the Orthocenter if AGC, with diameter AC.

Consequently, ∠AGC=90°

Typo: H is the circumcenter of AGC

DeletePalai Renato

ReplyDeleteBH/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

Reply

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

ReplyDeleteThen 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

Problem 1262

ReplyDeleteBH 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

Comment sent by Sumith Peiris:

ReplyDeleteDear 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

Thanks Sumith

DeleteSee problem 1263 at:

http://www.gogeometry.com/school-college/3/p1263-triangle-medians-90-degree-perpendicular.htm

@Sumith, thanks that makes the problem even more interesting

ReplyDeleteextend AC to AZ such that HB=HZ

ReplyDeletethen

LBZH=45.......so....LECH=45 etc.