April wrote:
The bisectors of trapezoid's angles form a quadrilateral with perpendicular diagonals. Prove that this trapezoid is isosceles.
$ \angle AHD=\angle BFC=90^o$. Let J,K be the midpoint of AD, BC
It's easy to see that $ HJ//AB, FK//AB$ but $ JK//AB$ so $ HF//AB$
Moreover $ EG\perp HF$ then $ \angle EHF=\angle EFH$
$ \Rightarrow \angle EDC=\angle ECD \Rightarrow$ QED
Sometime, your post seem quite unclearly to me .
Case1: $AD||BC$
In this case , the formed quadrilateral is a parallelogram and because its diagonals is perp with each other hence it's a rectangore
It makes our trapezoid is also a rectangle .
Done
Case2: $AD$ is not parallel to $BC$.WLOG : $AB<CD$
Let $AD$ meets $BC$ at J .
$I$ is $\triangle JAB$'s incenter
We have $IA||ED ; IB||EC \rightarrow E \in JI$
Similarly,we get :$HF||AB$
Hence :$HF \perp EG \Leftrightarrow IJ\perp AB \Leftrightarrow ABCD$ is isosceles