KOMATH

<Tri-2025-2280>, hint

<C'DA'=<C'AA'= 90-<B, A,D,C,A',C' are concyclic

<AA'C=<ADC=90.

<Tri-2025-2280> general, orthcenter

<Qua-2025-798>, hint

Let <ACD=X, <CAD=Y

<AFC=X+Y( <ABC+<AFC=180)=<ADF, AD=AF, Similarly CE=CD

Suppose Angle bisectors of <DAF , <DCE meet at O'

Above angles bisectors are also segments bisectors of DF, DE , which

meet at O , the circumcenter of DEF.

<CO'A=X+Y(By angle chasing)=<MON (M,N are the midpoints of DF, DE)

Hence O=O'.

<Qua-2025-798>, parallelogram, circimcenter

<Qua-2025-797>, hint

suppose a parallelogram ABCD and P,Q,R,S be the midpoint of side AB,BC,CD,DA

respectively. Since PS=QR=1/2 BD and for AQ=AR =a ,triangles ABQ and ADR be congruent,

Thus BQ=DR, say BC=AD

Hence ABCD should be a rhombus.