Geometry Problem. Post your solution in the comment box below.

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

Details: Click on the figure below.

## Sunday, May 26, 2019

### Geometry Problem 1436: Right Triangle, Altitude, Excircles, Excenters, Geometric Mean, Art, Poster

Labels:
altitude,
excenter,
excircle,
geometric mean,
geometry problem,
measurement,
right triangle

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connect AD, EC,

ReplyDeleteangle AHD =CHE =45 degree;

angle DAH =45+ angle ACB/2 ;

angle CEH = 180-45-angle ECH=135 - (45+angle BAC/2)= 90- angle BAC/2 =90 -(90-ACB)/2=45+angle ACB/2;

triangle ADH similar to CEH,

HD*HE=AH*HC=BH^2

https://photos.app.goo.gl/GRVbRNHhU7pWGEGf6

ReplyDeleteConnect AD and CE

Let a= ∠ (BAC)

In right triangle ABC we have ∠ (HBC)=a

Note that ∠ (EHC)= ∠ (AHD)=45

And ∠ (DAH)= (180-a)/2= 90-a/2

With angles manipulation we also get ∠ (HEC)= 90-a/2

So triangle AHD similar to EHC ( case AA)

And AH/EH= HD/HC => AH.HC= EH.HD

In right triangle ABC with altitude BH we have the relation BH^2=HA x HC= EH x HD