For reference: Show that the area of triangle $ABC = R\times MN(R=BO)$
I can't demonstrate this relationship
My progress:
$$S_{\triangle ABC} = \frac{abc}{4R}$$ $$S_{\triangle ABC}=\frac{AC\times BH}{2}$$
$BMHN$ is cyclic
Therefore $\angle HMN \cong\angle HBN\\ \angle MBH \cong \angle MNH\\$
$\triangle AMH \sim \triangle AHB\\ \triangle CNH \sim \triangle CHB$
...?
On
The orthocenter and the circumcenter of a triangle are isogonal conjugates, therefore
$\angle ABH=\angle NBO=α\\
HMBN(cyclic)\implies \angle BHM=\angle MNB=90−α\\
\therefore \angle BFN = 90^0 \implies BO\perp MN \\
[BMON]=[BMN]−[MNO]=\frac{BO⋅MN}{2}=\frac{R⋅MN}{2}(I)\\
\triangle MBN \sim \triangle CBA \therefore \frac{MN}{MB}=\frac{b}{a} \\
∠MHB=90−(90−∠A) \implies MB=BHsen(\angle A)\\
[BMON]=\frac{R⋅ba⋅BH⋅sen(\angle A)}{2}=\frac{[ABC]⋅Rsen(∠A)}{a}=\frac{[ABC]}{2}(II)\\
(I)=(II):\frac{[ABC]}{2}=\frac{R.MN}{2} \implies \boxed{[ABC]=R.MN}
$
(Solution by FelipeM.)
On
Hints for a geometric solution:
-Draw a circle radius $R$ center on B, it intersect altitude BH at E.
-Connect N to E and extend it to meet a circle center on N and radius MN at F. It can be seen that NF is parallel with AC, so it is perpendicular on BH at point E. So $FN=MN$. We have:
$S_{ABC}=S_{BFN}+S_{ACNF}=\frac{R\times FN}2+\frac{(FN+AC)(BH-R)}2$
which finally gives:
$R\times FN=FN\times BH-R\times FN + AC\times BH-AC\times R$
Or:
$2R\times FN=AC\times BH$
if you prove:
$AC\times R=FN\times BH$
First establish that $MN = BH \times \sin \angle ABC$
(law of sines - $\sin (\angle HMB):BH = \sin (\angle ABC):MN$).
Then,
$\triangle AOC$ is isosceles, let $\angle OAC=\angle OCA = \beta$.
Then $\angle AOC = 180^\circ - 2\beta = 180^\circ - \alpha-\gamma$.
where $\alpha,\gamma$ are the appropriate angles from the isosceles triangles $\triangle AOB$ and $\triangle BOC$.
The perpendicular from $O$ through $AC$ divides the angle in two, i.e. $90^\circ - \alpha-\gamma$, and so we arrive at $AC=R\times \sin (\angle ABC)$.