Proving orthocentre's existence

[Labib]

I'm such a WEAKLING that I couldn't solve this simple problem, but I bet you people can....

$PROBLEM::$

Prove that, cevians perpendicular to the opposite sides of a triangle are concurrent.

[tarek like math]

may be u mean indifferent lines ?

[*Mahi*]

I think you know the Trig Ceva , don't you? Use that to derive a more-than-easy proof.
[Trig Ceva :http://www.cut-the-knot.org/triangle/TrigCeva.shtml ]

[Labib]

Cevians are lines drawn from a vertex of a triangle to any point of it's opposite site. @tarek

@mahi, found a proof without trigCeva. but I'll try using trigCeva for a better one.

[photon]

trig ceva is cool..
in triangle $ABC$ let $AD,BE,CF$ ARE are perpendicular cevians.
in triangle $ABE$ and $ACF$
\[\angle A \]
is common
\[\angle AEB=\angle AFC=90\]
SO,
\[\angle ABE=\angle ACF\]
that's how we can show
\[\angle CBE=\angle DAC\]
\[\angle BCF=\angle BAD\]
that gives
\[\frac{sin\angle ABE }{sin\angle EBC}.\frac{sin\angle BCF}{sin\angle ACF}.\frac{sin\angle CAD}{sin\angle BAD}=1\]
orthocentre exists.
@Labib bhai,eager to see your proof without trigceva.

[tarek like math]

how it can prove without trig ceva ?

[Labib]

I'll give it now....

[Labib]

The proof's almost like yours!!


In triangle $\triangle ABC$ , let $AD,BE,CF$ are perpendicular cevians.
In triangles $\triangle ABE$ and $\triangle ACF$ ,
$\angle A$ is common,

$\angle AEB=\angle AFC=90 $ ,

SO,
$\angle ABE=\angle ACF $

that's how we can show...

$\angle CBE=\angle DAC $

$\angle BCF=\angle BAD $

From here, we can deduce that....

$\triangle ABE \sim \triangle ACF$

Therefore,

$AE/AF=BE/CF$

Thus,
$BF/BD=CF/AD$ and
$CD/CE=AD/BE$

So, $AE\cdot BF\cdot CD/AF\cdot BD\cdot CE=1$.

[photon]

hattrick proof!!!
in $ABC$ $AD,BE,CF$ altitudes and $DEF$ pedal triangle.now,
\[\angle FDA =\angle EDA......(1)\]
what we get from higher math geo theo 4.1
\[\angle BDA =\angle CDA=90.....(2)\]
$(2)-(1)$
\[\angle BDF =\angle CDE\]
THUS,\[\angle BFD =\angle AFE\]
\[\angle FEA =\angle DEC\]
here,\[\angle B=180-\angle BFD-\angle BDF=180-\angle AFE-\angle CDE
\]
\[OR,\angle B=180-(180-\angle A-\angle AEF)-(180-\angle C-\angle DEC)\]
\[OR,\angle B=180-\angle B-180+2\angle AEF\]
\[OR,2\angle B=2\angle AEF\]
\[OR,\angle B=\angle AEF\]
thus we can show,
\[\angle B=\angle AEF=\angle DEC\]
\[\angle C=\angle BFD=\angle AFE\]
\[\angle A=\angle BDF=\angle CDE\]
that's how we get triangle $BFD,DEC,AFE$ similar to $ABC$.
from $BFD,ABC$ similarity,
\[\frac{BF}{BD}=\frac{BC}{BA}\]
from $AFE,ABC$ similarity,
\[\frac{AE}{AF}=\frac{AB}{AC}\]
from $CDE,ABC$ similarity,
\[\frac{CD}{CE}=\frac{AC}{BC}\]
with the last three equation we get ceva applied.so orthocentre exits.

[Tahmid Hasan]

we know that the three perps bisect the angles of the orthic triangle and we also know that angle bisectors concur at one point(incenter).so the incenter of the orthic triangle in the orthocenter of the main triangle.