Let $AB$ and $CD$ be chords in a circle of center $O$ with $A , B , C , D$ distinct , and with the lines $AB$ and $CD$ meeting at a right angle at point $E$. Let also $M$ and $N$ be the midpoints of $AC$ and $BD$ respectively . If $MN \bot OE$ , prove that $AD \parallel BC$.
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johnkwon0328
04.02.2015 18:23
$AC<BD \implies OM>ON, EM<EN \implies OM^2+EN^2>ON^2+EM^2$ so it's contradiction. Same way for the opposite. Thus, $AC=BD \implies AD \parallel BC$
combidude
05.10.2015 22:55
johnkwon0328 wrote: $AC<BD \implies OM>ON, EM<EN \implies OM^2+EN^2>ON^2+EM^2$ so it's contradiction. Same way for the opposite. Thus, $AC=BD \implies AD \parallel BC$ I cannot understand your solution. Could you please explain elaborately?
Nymoldin
10.06.2020 18:49
What was the morality of this act?
Mahdi_Mashayekhi
31.12.2021 20:23
We will prove OND and OMA are congruent. note that both are similar to DEA. ∠OMA = 90 = ∠OND , OA = OD , ∠MOA = ∠COA/2 = ∠CBA = ∠EDA = ∠NDO ---> OND and OMA are congruent now we have DN = AM so DB = AC so AD || BC. we're Done.
NoobMaster_M
05.04.2023 00:02
By Brahmagupta's theorem, $ME \perp BD$ and $NE \perp AC$, Again $OM \perp AC$ and $ON \perp BD$ . So $ OM \parallel NE$ and $ ON \parallel ME $. So $OMEN$ is a parallelogram. As $MN \perp OE$ , so $OMEN$ is a rhombus. Now $\frac{AC}{2}=ME=NE=\frac{BD}{2}$. So $AC=BD$. By some angle chase we get $AD \parallel BC$