+1641 Cyclic quadrilateral's sides are AB = 11, BC = 4, CD = 12, and AD = 10. Find the length of diagonals AC and BD and prove that (AB*CD) + (BC*AD) = AC*BD.
+118687 I have shown it but I am sure I have not done it by the intended method
My method is far from elegant.
Find BD in terms of angle BCD using the cosine rule and also
Find BD in terms of angle BAD using the cosine rule
not that angle BCD and angle BAD are supplementary which means cos
Solve for angle
Then use to find distance BD
Do the same for AC
Then you can just substitute to show that they are the same.
I am sure this is not quite the intended solution.
Ask questions if you want to :)
+118687 like this
\(BD^2=144+16-96cos\angle BCD\\ BD^2=160-96cos\angle BCD\\~\\ but\;\;also\;\; BD^2=121+100-220cos \angle BAD\\ BD^2=221-220cos \angle BAD\\ BD^2=221+220cos \angle BCD\qquad \text{opposite angles in cyclic quads are supplementary}\\~\\ so\\ 160-96cos\angle BCD=221+220cos \angle BCD\\ -316cos\angle BCD=61\\ cos\angle BCD=\frac{-61}{316}\\ so\\ BD^2=160-96*\frac{-61}{316}\\ BD^2=\frac{14104}{79}\\ BD=\sqrt{\frac{14104}{79}}\\ \)
Now do the same for AC
etc
LaTex:
BD^2=144+16-96cos\angle BCD\\
BD^2=160-96cos\angle BCD\\~\\
but\;\;also\;\;
BD^2=121+100-220cos \angle BAD\\
BD^2=221-220cos \angle BAD\\
BD^2=221+220cos \angle BCD\qquad \text{opposite angles in cyclic quads are supplementary}\\~\\
so\\
160-96cos\angle BCD=221+220cos \angle BCD\\
-316cos\angle BCD=61\\
cos\angle BCD=\frac{-61}{316}\\
so\\
BD^2=160-96*\frac{-61}{316}\\
BD^2=\frac{14104}{79}\\
BD=\sqrt{\frac{14104}{79}}\\
+118687 Thanks Dragen 
I really do think that there must be a better way to tackle this problem though.
+1490 There's more than one way to... 
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We can use this formula to calculate the circumradius. (R-radius, s-semiperimeter, a,b,c,d-sides)
When we have the length of a circumradius, we can calculate the angles B and C or A and D, and then using the law of cosines we can find diagonals AC and BD.
This is perhaps the longer way, but it would work.
+118687 Mmm
You are using terms and formula that I have not seen before.
But yes, this is probably another way to skin the cat. (sorry feline lovers - it is just an expression)
+2094 Hi Melody. Maybe you can use some other phrases that do not put cats in such a situation, even if it's an expression.
The neighborhood cat Winston is bristling besides me.
AB.CD + BC.AD = AC.BD
is known as Ptolemy's theorem,
dates back around two thousand years.
