Some speciality of ortho and right angles!

Geometry Level 5

Let A B C ABC be a triangle with A B = 13 , B C = 14 AB=13, BC=14 and C A = 15. CA=15. Let H H be the orthocenter of △ A B C . \triangle ABC. Find the distance between the circumcenters of triangles A H B AHB and A H C . AHC.


The answer is 14.

Ayush G Rai by Ayush G Rai , 20, India

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3 solutions

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Solution without words

Vishwash Kumar ΓΞΩ - 4 years, 6 months ago

Great sir Camfar

Himanshu Srivastava - 4 years, 5 months ago

Thanks for your kind words.

Vishwash Kumar ΓΞΩ - 4 years, 5 months ago

Let D be the foot of altitude from A, and E foot of altitude from B..
ABD is a rt. triangle 13-5-12. Also ADC is a rt. triangle 12-9-15. So AD=12, BD=5, DC=9.
Using coordinate geometry we have,
A(0,12), B(-5,0), C(9,0) and D(0,0).
Slope of AC= -AD/DC= -4/3. So Slope of BE= - 1/( - 4/3)=3/4.
So line BE is Y=3/4(X+5)+0=3/4X+15/4.
But H is on BE as well on AD with equation X=0.
So H(0,15/4).
From the coordinates of vertices of ABH and AHC,
we calculate the coordinates of O,circumcenter of ABH as O( - 7, 63/8),
and the coordinates of O',circumcenter of AHC as O'( 7, 63/8).
Clearly OO' =7-(-7)=14.



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