Deriving the Distance Between Orthocentre & Circumcentre of a Triangle

[zorro]

The problem
Can anyone explain me a step given in my book?

This is actually the derivation of the distance between orthocentre and circumcentre of a triangle using vectors.

ABC is a triangle.
S is the circumcentre and H is the orthocentre.
$$\vec{SA}$$ + $$\vec{SB}$$ + $$\vec{SC}$$ = $$\vec{SA}$$ + 2$$\vec{SD}$$
=$$\vec{SA}$$ + $$\vec{AH}$$

I just want to find out how 2$$\vec{SD}$$ = $$\vec{AH}$$

 

[chaoseverlasting]

What is D?

 

[zorro]

OH! I'm sorry.
D is the midpoint of BC

 

[chaoseverlasting]

Thats because AD is a median from A to BC. The circumcenter divides the median into a ratio of 2:1. Thus AH=2SD.
Btw, how did the derivation come to SC+SB=2SD? The way I got that result is that BSC is a triangle that's half of a parallelogram. Since D is the midpoint of BC, its the point of intersection of the diagonals. Which means the other diagonal is of length 2SD. The vector sum of the two sides, thus, must be equal to the diagonal. Is there another way to do this?

 

[zorro]

Thats because AD is a median from A to BC. The circumcenter divides the median into a ratio of 2:1. Thus AH=2SD.

I don't think that circumcenter divides the median in the ratio 2:1.
It is the centroid which dividies the median in that ratio.

Moreover, H is the orthocentre, and it does not lie on AD.

Btw, how did the derivation come to SC+SB=2SD? The way I got that result is that BSC is a triangle that's half of a parallelogram. Since D is the midpoint of BC, its the point of intersection of the diagonals. Which means the other diagonal is of length 2SD. The vector sum of the two sides, thus, must be equal to the diagonal. Is there another way to do this?

That's right. Maybe there is another method, but this one is the shortest.

 

[zorro]

I got the answer.
We have to use the fact that the centroid divides orthocentre and circumcentre in the ratio 2:1

 

[chaoseverlasting]

The centeroid is the point of intersection of medians. In this case the circumcenter and the median are coincide. The medians are lines from a vertex to the midpoint of the opposite line of a triangle. Since in this case the lines are perpendicular to the other side of the triangle, this point is the circumcenter as well as the centeroid. Thus, it divides the median in a 2:1 ratio.

I got the answer.
We have to use the fact that the centroid divides orthocentre and circumcentre in the ratio 2:1

I may be wrong, but this sounds suspect. In this case, the centeroid and the circumcenter coincide. There can be no line division.

 

[zorro]

You are wrong. Except in an equliateral triangle, circumcentre and centroid don't coincide.
Here it is just given that D is the midpoint of BC. S is the circumcentre. It is not necessary that SD should coincide with AD ( the median ).

 

[chaoseverlasting]

You're right, I made a mistake. Sorry.

Could you post your solution if possible? I've come about it in a very round about manner.

EDIT: Got it.