BdMO National 2013: Secondary 2, Higher Secondary 1

BdMO
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BdMO National 2013: Secondary 2, Higher Secondary 1

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.

Fatin Farhan
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Re: BdMO National 2013: Secondary 2, Higher Secondary 1

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sowmitra
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Re: BdMO National 2013: Secondary 2, Higher Secondary 1

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.
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Pentagons- Not So Degenerate.png (45.23 KiB) Viewed 2459 times
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Re: BdMO National 2013: Secondary 2, Higher Secondary 1

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.