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#1 h Apr 4, 2007, 9:02 PM • 1 Y

Y by Adventure10

$E$ and $F$ are points on sides $AB$ and $BC$ of square $ABCD$ respectively such that: $BE=BF$ . If $BN$ is the altitude of triangle $BEC$ , Prove that $DNF$ is right angle.

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#2 h Apr 4, 2007, 10:30 PM • 2 Y

Y by Adventure10, Mango247

We denote as $M,$ the intersection point of $BD,$ $EF$ and so, we have that the quadrilateral $BENM$ is cyclic, because of $BD\perp EF.$

Hence, we conclude that $\angle NEF = \angle NBD$ $,(1)$

From the similar right triangles $\bigtriangleup BNE,$ $\bigtriangleup BNC,$ we have that

$\frac{EN}{BN}= \frac{BE}{BC}=\frac{EF}{AC}= \frac{EF}{BD}$ $\Longrightarrow$ $\frac{EN}{BN}= \frac{EF}{BD}$ $,(2)$

From $(1),$ $(2)$ we conclude that the triangles $\bigtriangleup ENF,$ $\bigtriangleup BND,$ are similar and so, we have that $\angle NDB = \angle NFE$ $,(3)$

From $(3),$ $\Longrightarrow$ the quadrilateral $DNMF,$ is cyclic.

Hence, we have that $\angle DNF = \angle DMF = 90^{o}$ and the pproof is completed.

Kostas vittas.

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#3 h Apr 5, 2007, 8:06 AM • 1 Y

Y by Adventure10

sinajackson wrote:

$E$ and $F$ are points on sides $AB$ and $BC$ of square $ABCD$ respectively such that: $BE=BF$ . If $BN$ is the altitude of triangle $BEC$ , Prove that $DNF$ is right angle.

It is a nice problem from Leningrad or All Russian olympiad..

After Vitta's nice (and new to me !) solution , I send you one more solution, not metric.

If $BN$ meets $AD$ at $Z$ , then triangles $ABZ, BEC$ are equal (since $\angle ABZ= \angle BCE$ . Thus $AZ=BE=BF$ and $DZ=CF$ . If $DF$ meets $CZ$ at $O$ , then :
$ON= \frac{ZC}{2}= \frac{DF}{2}$ , because in right triangle $ZNC$ segment $NO$ is median and $ZC=DF$ .

This means that triangle $DNF$ is right , since $NO$ is a median in this triangle.

This post has been edited 1 time. Last edited by stergiu, Apr 5, 2007, 8:14 AM

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#4 h Apr 5, 2007, 8:10 AM • 1 Y

Y by Adventure10

A nice and easy problem !A short proof.Stergiu, are only four minutes between the our messages !

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#5 h Apr 5, 2007, 8:45 AM • 1 Y

Y by Adventure10

Virgil Nicula wrote:

Stergiu[/u][/b], are only four minutes between the our messages !

It seems , we solve problems about the same time !!!

Greetings.-

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#6 h Apr 5, 2007, 8:46 AM • 1 Y

Y by Adventure10

A metrical proof with the following simple remark applied to the right triangle NBC :Indeed, $\tan\widehat{CNF}=\frac{CF}{CB}=\frac{CF}{CD}=\tan\widehat{CDF}$ , i.e. $\widehat{CNF}\equiv\widehat{CDF}$ a.s.o.

Stergiu, (again) now is only one minute between our messages ! Greetings !

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