BDMO NATIONAL Junior 2016/04

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BDMO NATIONAL Junior 2016/04
please give the solution in picture .i can't understand the writting
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 ahmedittihad
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Re: BDMO NATIONAL Junior 2016/04
This is known as the perpendicular lemma. It is quite handy in proving perpendicularity.
PROOF
Let $AC\cap BD=E$. Now, apply the Law of Cosines on triangles $ABE$,$CBE$,$CDE$,$DAE$. Let $\angle AEB=\theta$. We have
$$\begin{aligned}
BA^2&=AE^2+BE^22(AE)(BE)(\cos \theta)\\
BC^2&=CE^2+BE^2+2(BE)(CE)(\cos \theta)\\
DC^2&=DE^2+CE^22(DE)(CE)(\cos \theta)\\
AD^2&=AE^2+DE^2+2(AE)(DE)(\cos \theta)\\
\end{aligned}$$
and substituting into $AB^2AD^2=BC^2CD^2$ yields that $\cos\theta=0$. So, $\theta=90$.
PROOF
Let $AC\cap BD=E$. Now, apply the Law of Cosines on triangles $ABE$,$CBE$,$CDE$,$DAE$. Let $\angle AEB=\theta$. We have
$$\begin{aligned}
BA^2&=AE^2+BE^22(AE)(BE)(\cos \theta)\\
BC^2&=CE^2+BE^2+2(BE)(CE)(\cos \theta)\\
DC^2&=DE^2+CE^22(DE)(CE)(\cos \theta)\\
AD^2&=AE^2+DE^2+2(AE)(DE)(\cos \theta)\\
\end{aligned}$$
and substituting into $AB^2AD^2=BC^2CD^2$ yields that $\cos\theta=0$. So, $\theta=90$.
Frankly, my dear, I don't give a damn.

 Posts: 29
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 Location: Rajshahi,Bangladesh
Re: BDMO NATIONAL Junior 2016/04
please, give the answer normal geometry
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: BDMO NATIONAL Junior 2016/04
I'm afraid the other solution I know is with vectors.
Frankly, my dear, I don't give a damn.
Re: BDMO NATIONAL Junior 2016/04
Let's assume $AC$ is not perpendicular to $BD$.
WLOG, lets assume $\angle AOB < 90$. Let $X$ be the foot of the perpendicular from $B$ to AC and let $Y$ be the foot of the perpendicular from $D$ to $AC$.
Now, extension of Pythagporous theorem gives
$BO^2 + AO^2 + 2.AO.OX + DO^2 + CO^2+ 2.CO.OY = BO^2 + OC^2  2.OC.OX + DO^2 + AO^2  2AO.OY$
Or, $2.AO.OX + 2.CO.OY = (2.OC.OX+2.AO.OY)$
Or, $2AO(OX+OY) = 2.OC(OX+OY)$
But that is only possible when $OX = OY = 0$, or $X,Y$ lies on $O$. So, the diagonals $AC,BC$ are perpendicular to each other.
WLOG, lets assume $\angle AOB < 90$. Let $X$ be the foot of the perpendicular from $B$ to AC and let $Y$ be the foot of the perpendicular from $D$ to $AC$.
Now, extension of Pythagporous theorem gives
$BO^2 + AO^2 + 2.AO.OX + DO^2 + CO^2+ 2.CO.OY = BO^2 + OC^2  2.OC.OX + DO^2 + AO^2  2AO.OY$
Or, $2.AO.OX + 2.CO.OY = (2.OC.OX+2.AO.OY)$
Or, $2AO(OX+OY) = 2.OC(OX+OY)$
But that is only possible when $OX = OY = 0$, or $X,Y$ lies on $O$. So, the diagonals $AC,BC$ are perpendicular to each other.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
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 Charles Caleb Colton

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Re: BDMO NATIONAL Junior 2016/04
plz say what is perpendicullar lemma. i am protaya das. a winner of national bdmo on 2017 from junior category
 ahmedittihad
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 Joined: Mon Mar 28, 2016 6:21 pm
Re: BDMO NATIONAL Junior 2016/04
Hello Protya Das, winner of bdmo 2017, the perpendicular lemma is thisprotaya das wrote: ↑Sun Dec 10, 2017 3:34 pmplz say what is perpendicullar lemma. i am protaya das. a winner of national bdmo on 2017 from junior category
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Frankly, my dear, I don't give a damn.

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Re: BDMO NATIONAL Junior 2016/04
hello protaya and prottoy das are the same person.I have forgot the password of protaya das so i am using the prottoy das id.

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Re: BDMO NATIONAL Junior 2016/04
the problem is from the book euclidean geometry in mathmatical olympiad