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BdMO National 2013: Secondary 2, Higher Secondary 1
Posted: Fri Jan 10, 2014 1:33 am
by BdMO
A polygon is called degenerate if one of its vertices falls on a line that joins its neighboring two vertices. In a pentagon $ABCDE$, $AB=AE$, $BC=DE$, $P$ and $Q$ are midpoints of $AE$ and $AB$ respectively. $PQ||CD$, $BD$ is perpendicular to both $AB$ and $DE$. Prove that $ABCDE$ is a degenerate pentagon.
Re: BdMO National 2013: Secondary 2, Higher Secondary 1
Posted: Sun Jan 12, 2014 5:32 pm
by Fatin Farhan
Re: BdMO National 2013: Secondary 2, Higher Secondary 1
Posted: Sun Jan 12, 2014 6:21 pm
by sowmitra
Fatin Farhan wrote:In the quadrilateral $$BCDE$$, $$CD||BE, \angle ABE=\angle BED$$.
So $$BCDE$$ is a parallelogram.
$\angle ABE=\angle BED$ does not necessarily imply that $BCDE$ is a parallelogram. Actually, $BCDE$ can be a Parallelogram,
OR, an Isosceles-Trapezium.
**Note: This is actually a hole in the problem, as there aren't any additional conditions to determine whether $BCDE$ is a Parallelogram or an Isosceles Trapezium. So, the given construction doesn't always produce a degenerate pentagon.
As you can see below, both the pentagons $ABCDE$ and $ABC'DE$ satisfy all the given conditions. But, $ABCDE$ is non-degenerate, whereas, $ABC'DE$ is degenerate.
Re: BdMO National 2013: Secondary 2, Higher Secondary 1
Posted: Sun Jan 12, 2014 6:37 pm
by *Mahi*
sowmitra wrote:
**Note: This is actually a hole in the problem, as there aren't any additional conditions to determine whether $BCDE$ is a Parallelogram or an Isosceles Trapezium. So, the given construction doesn't always produce a degenerate pentagon.
Yep, there actually is one. This question statement was false, even if it was given as a no. 1 in higher secondary.
Re: BdMO National 2013: Secondary 2, Higher Secondary 1
Posted: Sat Dec 24, 2022 11:10 pm
by A.K.M.Zakaria
AB=AC
AQ=AP [P,Q are the midpoints of AE & AB]
.'.PQ||BE
But, PQ||CD
.'.BE||CD