## BDMO REGIONAL 2015

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Posts: 29
Joined: Mon Jan 23, 2017 10:32 am

### BDMO REGIONAL 2015

Hey, guys .Help me to solve this.....
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Absur Khan Siam
Posts: 65
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

### Re: BDMO REGIONAL 2015

Let's draw a line such that it goes through $E$ and
perpendicular to $AB$ and $CD$.It intersects $AB$
at $F$ and $CD$ at $G$.We also can say that $BF = CG , AF = DG$.
We can form four equations:
$DE^2 = EG^2 + DG^2...(i)$
$CE^2 = EG^2 + CG^2...(ii)$
$AE^2 = EF^2 + AF^2...(iii)$
$BE^2 = EF^2 + BF^2...(iv)$
$(iv) -(iii) \Rightarrow BE^2 - AE^2 = BF^2 - AF^2 \Rightarrow 20 = BF^2 - AF^2$
$= CG^2 - DG^2$
$(i) - (ii) \Rightarrow DE^2 = DG^2 - CG^2 + CE^2 = -20 + 25 = 5$
$\therefore DE = \sqrt{5}$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

Absur Khan Siam
Posts: 65
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

### Re: BDMO REGIONAL 2015

A simpler solution:
Draw two diagonals $AC$ and $BD$ and let them intersect in $O$.
$AO = BO = CO = DO$
Thus,in $\triangle AEC$ , $EO$ is the median.
Appolonius' theorem implies that $AE^2 + CE^2 = 2(AO^2 + EO^2)$
Similarly , from $\triangle BED$ , $DE^2 + BE^2 = 2(BO^2 + EO^2) = 2(AO^2 + EO^2)$

Thus , we can say ,$AE^2 + CE^2 = BE^2 + DE^2 \Rightarrow DE^2 = AE^2 + CE^2 - BE^2 = 5$

$\therefore DE = \sqrt{5}$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

samiul_samin
Posts: 1004
Joined: Sat Dec 09, 2017 1:32 pm

### Re: BDMO REGIONAL 2015

This is BdMO regional Mymensingh higher secondary problem no $6$.