BdMO National Higher Secondary 2008/8

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
User avatar
Moon
Site Admin
Posts:751
Joined:Tue Nov 02, 2010 7:52 pm
Location:Dhaka, Bangladesh
Contact:
BdMO National Higher Secondary 2008/8

Unread post by Moon » Sun Feb 06, 2011 11:19 pm

Problem 8:
$ABCD$ is a cyclic quadrilateral. The diagonals $AC$ and $BD$ intersect at $E$.$ AB = 39; AE = 45; AD = 60; BC = 56$. Find the length of $CD$.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

User avatar
bristy1588
Posts:92
Joined:Sun Jun 19, 2011 10:31 am

Re: BdMO National Higher Secondary 2008/8

Unread post by bristy1588 » Sat Jan 14, 2012 10:42 am

Is the answer
$18.2$?
Bristy Sikder

samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm

Re: BdMO National Higher Secondary 2008/8

Unread post by samiul_samin » Wed Feb 21, 2018 11:54 pm

Moon wrote:
Sun Feb 06, 2011 11:19 pm
Problem 8:
$ABCD$ is a cyclic quadrilateral. The diagonals $AC$ and $BD$ intersect at $E$.$ AB = 39; AE = 45; AD = 60; BC = 56$. Find the length of $CD$.
How to solve this problem?

Ahmed Ashhab Mahir
Posts:2
Joined:Mon Aug 16, 2021 9:10 pm

Re: BdMO National Higher Secondary 2008/8

Unread post by Ahmed Ashhab Mahir » Fri Apr 22, 2022 1:51 am

Let $\angle{CAB}=x$ $\angle{ABD}=f$ $\angle{CAD}=y$ $\angle{ADB}=z$

As $ABCD$ a cyclic quadrilateral . So $ \frac{56}{sin(x)}=\frac{60}{sin(f)}=\frac{39}{sin(z)}=\frac{CD}{sin(y)}=2R$

Now in $\triangle ABE$,
$\frac{sin(f)}{45}=\frac{sin(x)}{BE}$
So, We get
$BE=42$

Now in $\triangle ADE$

$\frac{sin(z)}{45}=\frac{sin(y)}{DE}$

So $\frac{39}{45}=\frac{CD}{DE} Or, CD=\frac{39.DE}{45}$

By intersecting chord theorem,
$DE.BE=AE.CE$

$\frac{DE}{CE}=\frac{AE}{BE}=\frac{45}{42}$

$DE=\frac{CE.45}{42}$

So $CD=\frac{39.CE}{42}$
By Steward's theorem , $AB^2.CE+BC^2.AE=AC(BE^2+CE.AE)$

$39^2.CE+56^2.45=(45+CE)(42^2+45CE)$

Solve for CE, you will get $CE=19.6$
Putting CE in $CD=\frac{39.CE}{42}$ , We get $CD=\fbox{18.2}$

lilywatson1v
Posts:1
Joined:Tue May 10, 2022 8:26 am

Re: BdMO National Higher Secondary 2008/8

Unread post by lilywatson1v » Tue May 10, 2022 8:29 am

Thank you for bringing new knowledge to everyone, it is very helpful
1v1 battle build now gg

Post Reply