Blanchet's Theorem

by henderson, Oct 3, 2016, 1:05 PM

$\color{blue}\bf{Blanchet's \ theorem }$ In an acute $\triangle ABC,$ $F$ is the foot of the perpendicular from $C$ to $AB$ and $P$ is a point on $CF.$ Let the lines $AP$ and $BP$ meet $BC$ and $AC$ at $D$ and $E,$ respectively. Show that $\angle EFC = \angle DFC.$

Blanchets Theorem

geometry

Comment

4 Comments

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Yeah,it is very useful and important theorem.

by Orkhan-Ashraf_2002, Nov 17, 2016, 6:44 PM

Report

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Let $DE\cap AB\equiv X,DE\cap CF\equiv Y\implies (X,Y;D,E)=(-1)$ , but $FY\perp FX$ , so $FY$ is the internal angle bisector of $\angle DFE\implies \angle EFC=\angle DFC$ .

by babu2001, Jul 3, 2017, 12:15 PM

Report

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

as the figure shows, PQ is the line parallel to AB and through point C .thus,we only have to prove that QC=PC
due to the two parallel lines ,we can work out that QC=AF*EC/AE PC=BF*CD/BD
CEVA theorem tells us that AF*EC*BD=BF*CD*AE so QC=PC proved.

Attachments: