Question

Let $H$ be the orthocenter of the equilateral triangle $\triangle ABC$. We know the distance between the orthocenters of $\triangle AHC$ and $\triangle BHC$ is $12$. What is the distance between the circumcenters of $\triangle AHC$ and $\triangle BHC$?

Answer 1

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Step-by-step explanation:

Answer 2

The distance between the circumcenters of Triangle AHC and Triangle BHC is 6.

Given that,

Let H be the orthocenter of the equilateral triangle.

The distance between the orthocenters of triangle AHC and triangle BHC is 12.

We have to find,

What is the distance between the circumcenters of Triangle AHC and Triangle BHC?

According to the question,

Here, H is the Orthocenter of the equilateral triangle.

And the distance between the orthocenters of triangle AHC and triangle BHC is 12.

By the relation between the orthocenter and circumcenter are always collinear.

Centroid divided the line joining orthocenter and circumcenter in the ratio 2;1.

Therefore,

The distance between the circumcenters of Triangle AHC and Triangle BHC is,

$\rm Circumcenter = \dfrac{1}{2} \times Orthocenter\\\\ Circumcenter = \dfrac{1}{2} \times 12 \\\\ Circumcenter = 1 \times 6 \\\\ Circumcenter = 6$

Hence, The distance between the circumcenters of Triangle AHC and Triangle BHC is 6.

For more details refer to the link given below.

https://brainly.com/question/19763099