geomerty

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geomerty
plz, help me solving this
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Re: geomerty
Let's extend AC and define two points $F$ and $E$ such that $CE = BC$ and
$CD = CF$.
By some angle chasing,
$\angle DEF = \angle EDF \Rightarrow DF = EF$
$\angle DFA = \angle FAD \Rightarrow DF = AD$
$\therefore AD = EF$
Now ,
$AD + CD = EF + CF = CE = BC = 2017$
$CD = CF$.
By some angle chasing,
$\angle DEF = \angle EDF \Rightarrow DF = EF$
$\angle DFA = \angle FAD \Rightarrow DF = AD$
$\therefore AD = EF$
Now ,
$AD + CD = EF + CF = CE = BC = 2017$
"(To Ptolemy I) There is no 'royal road' to geometry."  Euclid
Re: geomerty
Summation of any two side of a triangle is bigger than the third side.Absur Khan Siam wrote:Let's extend AC and define two points $F$ and $E$ such that $CE = BC$ and
$CD = CF$.
By some angle chasing,
$\angle DEF = \angle EDF \Rightarrow DF = EF$
$\angle DFA = \angle FAD \Rightarrow DF = AD$
$\therefore AD = EF$
Now ,
$AD + CD = EF + CF = CE = BC = 2017$
But u have showed that,$AD + CD = EF + CF = CE = BC = 2017$,Here $$AD+CD=BC$$ which is practically impossible.?!

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 Joined: Sun Dec 11, 2016 2:01 pm
Re: geomerty
<FAD itself measures 100 degree,then how can you concurr that AD=DF?I think you did not consider triangle DAF. ..According to your construction,<BCE=140 as it is supplementary to <ACB. Therefore,as CF=CD, <FCD=160 and <DFC=<DFA merely measures 10...check out your angle chasing...
 Thamim Zahin
 Posts: 98
 Joined: Wed Aug 03, 2016 5:42 pm
Re: geomerty
This is a Dhaka regional 2017. I solved it using scale and compass(just measured the length with my scale. Got $2017$, which was definitely the answer because this is 2017) :p
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Re: geomerty
I am showing the angle chasing part
Extend $AC$ in such a way so that $F,E$ are on the opposite side of $C$. Now, $\angle CFD = \angle CDF = 80$
$\angle FDA = 20$, $\angle DAF = \angle DFA = 80$, so, $DA = DF$.
Again, $\angle CBE = \angle CEB = 70$, $\angle DBE = 30$. Now, extend $CD$ so that it meets $BE$ at $X$. $\angle CXE = 90$.Now, $\bigtriangleup CXE $ is congruent to $\bigtriangleup CXB$. So, $BX = XE$. SO, $\bigtriangleup BDE$ is isosceles and $\angle DEB = 30 \Rightarrow \angle DEF = 40$. So, $\bigtriangleup DEF$ is isosceles and $DF = FE$. So, $CD + AD = CF + DF = CF + EF = CE = BC = 2017$.
Extend $AC$ in such a way so that $F,E$ are on the opposite side of $C$. Now, $\angle CFD = \angle CDF = 80$
$\angle FDA = 20$, $\angle DAF = \angle DFA = 80$, so, $DA = DF$.
Again, $\angle CBE = \angle CEB = 70$, $\angle DBE = 30$. Now, extend $CD$ so that it meets $BE$ at $X$. $\angle CXE = 90$.Now, $\bigtriangleup CXE $ is congruent to $\bigtriangleup CXB$. So, $BX = XE$. SO, $\bigtriangleup BDE$ is isosceles and $\angle DEB = 30 \Rightarrow \angle DEF = 40$. So, $\bigtriangleup DEF$ is isosceles and $DF = FE$. So, $CD + AD = CF + DF = CF + EF = CE = BC = 2017$.
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 Charles Caleb Colton
 Charles Caleb Colton
 Thanic Nur Samin
 Posts: 176
 Joined: Sun Dec 01, 2013 11:02 am
Re: geomerty
Here is an easy solution:
Take $K$ on $BC$ so that $CD=CK$. Now, since $\triangle CDK$ is isosceles and $\angle DCK=20^{\circ}$, $\angle DKC=80^{\circ}$ implying $ADKC$ is cyclic.
Now, $\angle BDK=\angle BCA=\angle ABC$, so $BK=DK$. Again, $CD$ is the angle biscector, so $AD=DK$. So $AD+CD=BK+KC=BC=2017$
Take $K$ on $BC$ so that $CD=CK$. Now, since $\triangle CDK$ is isosceles and $\angle DCK=20^{\circ}$, $\angle DKC=80^{\circ}$ implying $ADKC$ is cyclic.
Now, $\angle BDK=\angle BCA=\angle ABC$, so $BK=DK$. Again, $CD$ is the angle biscector, so $AD=DK$. So $AD+CD=BK+KC=BC=2017$
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 Atonu Roy Chowdhury
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 Location: Chittagong, Bangladesh
Re: geomerty
The simplest solution seems to me:
1. Use Cosine Law to find $AB$
2. Angle Bisector Theorem to find $AD$
3. Cosine Law on $\triangle ADC$ to find $CD$
1. Use Cosine Law to find $AB$
2. Angle Bisector Theorem to find $AD$
3. Cosine Law on $\triangle ADC$ to find $CD$
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 ahmedittihad
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 Atonu Roy Chowdhury
 Posts: 63
 Joined: Fri Aug 05, 2016 7:57 pm
 Location: Chittagong, Bangladesh
Re: geomerty
In $\triangle ABC$ , $c^2 = a^2 + b^2  2ab \cos C$ahmedittihad wrote:What is cosine law?
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