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In the following figure prove that: $OA$ is perpendicular to $B'C'$ ($ABC$ is an arbitrary triangle,$BB'$ and $CC'$ are heights and $O$ is circumcenter) enter image description here

I noticed that $B,C',B',C$ are concyclic, but how does this fact help?(another set of 4 concyclic points is also seen in the figure)

Hamid Reza Ebrahimi
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    You may use the fact that $O$ and $H$ are isogonal conjugates and $BCB'C'$ is a cyclic quadrilateral. Since $AH\perp BC$, $AO\perp B'C'$ follows. – Jack D'Aurizio Sep 27 '16 at 09:12

3 Answers3

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Construction:- 1) Join OB; 2) Drop a perpendicular from O to AB.

enter image description here

$\angle 1 = \angle 2$ [congruent triangles]

$\angle 1 + \angle 2 = 2 \times \angle 3$ [Angle at center = 2 times angle at circumference]

$\angle 3 = \angle 4$ [exterior angle of cyclic quad.].

Result follows.

Added: enter image description here

Mick
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assume the circle centered at the origin $O(0, 0)$ & the coordinates $A(x_1, y_1)$, $B(x_2, y_2)$ & $C(x_3, y_3)$ then find the coordinates of foots of perpendiculars $B'$ & $C'$ & then find out slope of line B'C' & line OA then show the product of slope to be $-1$

Bhaskara-III
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Let me give you a geometrical proof:

Let J be the point where AO intersects C'B'

Assume B'BC=x

Then B'CB=90-x and hence AOB=180-2x (why?)

B'C'C= x

Hence B'C'A=90-X (why?)

Look at ΔAOB : AO=OB hence it is isosceles and then we have BAO=x.

Now look at ΔAJC where C'AO=x and AC'B' = 90-x.

Thus AJC=90 .

  • I hope you can figure out the reasons behind some steps since they follow from basic geometrical lemmas. –  Sep 27 '16 at 09:08