BdMO National Higher Secondary 2008/8

Moon
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BdMO National Higher Secondary 2008/8
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 $CD$.
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bristy1588
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Re: BdMO National Higher Secondary 2008/8

$18.2$?
Bristy Sikder

samiul_samin
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Re: BdMO National Higher Secondary 2008/8

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

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
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Joined:Tue May 10, 2022 8:26 am

Re: BdMO National Higher Secondary 2008/8

Thank you for bringing new knowledge to everyone, it is very helpful