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secondary dhaka divisional 2015
Posted: Wed Dec 28, 2016 2:55 am
by Nodee Haque
$ABCD$ is a parallelogram and it's diagonals meet at point $O$. $P$ an $Q$ are the midpoints of $AO$ and $BC$ consecutively. $\angle A = \angle DPQ$ and $\angle DBA = \angle DPQ$. If $AB = 2$ unit, then find out the area of $ABCD$.
Re: secondary dhaka divisional 2015
Posted: Wed Jan 04, 2017 3:36 am
by ahmedittihad
Interestingly, this problem needs spiral similarity.
So, we have, $\triangle DAB$ similar to $ \triangle DPQ $.
By spiral similarity, $\triangle DQB $ similar to $ \triangle DPA $.
So, $\angle DAP=\angle DBQ$. This is only possible when the parallelogram is a rectangle.
We also have, for similarity reasons, $DA/AP = DB/BQ$ or, $DA/AO = DB/BC$.
Along with the condition that, $\angle DAP=\angle DBQ$ we get that $\triangle AOD $ similar to $\triangle BCD$. We get that, $\angle AOD =90$. So, we get, $ABCD$ is a square. So the answer is $4$.