$ABCD$ is a parallelogram and $O$ is any point. The parallelograms $OAEB$, $OBFC$, $OCGD$, $ODHA$ are completed. Show that $EFGH$ is a parallelogram
I did see this question but it doesn't have the complete answer.
The main problem is to prove the co-linearity of the points $EBF, GCF, HDG$, and $EAH$, which I am unable to figure out. After proving that, the question becomes straight forward.
To prove $\angle EBF=180^\circ$,
$180^\circ - \angle ABC = \angle DCB$ and $\angle CBF = \angle OCB$
If we prove $\angle DCO= \angle ABE$, the question can be solved. Is there any other method to solve this question? Thanks in advance
I can send you the figure uf you give me an address.
Hint: In figure I constructed following points are considered, the intersection of diagonals of ABCD is M.
1- In parallelogram OAEB , vertex E is mirror of O about the midpoint of AB.Also $AE||OB||FC$, so we can construct parallelogram AECF with diagonals AC and EF which passes point M.
2- In parallelogram ODGC , vertex G is mirror of O about the midpoint of DC.also $ GC||OD||AH$, so we can construct parallelogram AGCH with diagonals AC and HG. The midpoint of this diagonal is coincident on intersection of diagonals of ABCD, i.e M.
3- In parallelogram ODAH , vertex H is such that $HD||AD||BC$.
4- In parallelogram OBFC , vertex F is such that $OF||BC$, also $FC||OB||AE$.
From 3 and 4 we conclude that OH and OF are co-linear.
In triangles AHE and CGF two sides are equal and parallel i.e $GC(||=)AH$ and $FC(||=)AE$, that is they are equal and their third sides are also parallel, ie, $HE||FG$.So EF can also be the diagonal of a parallelogram as well as it is the diagonal of parallelogram AECF mentioned in 1. That is $EG||HF$ and EFGH is a parallelogram.