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BdMO National Higher Secondary 2008/8
Posted: Sun Feb 06, 2011 11:19 pm
by Moon
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 $CD$.
Re: BdMO National Higher Secondary 2008/8
Posted: Sat Jan 14, 2012 10:42 am
by bristy1588
Is the answer
$18.2$?
Re: BdMO National Higher Secondary 2008/8
Posted: Wed Feb 21, 2018 11:54 pm
by samiul_samin
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?
Re: BdMO National Higher Secondary 2008/8
Posted: Fri Apr 22, 2022 1:51 am
by Ahmed Ashhab Mahir
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 $CD=\fbox{18.2}$
Re: BdMO National Higher Secondary 2008/8
Posted: Tue May 10, 2022 8:29 am
by lilywatson1v
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