+0  
 
+2
123
15
avatar+1164 

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.

 

 Mar 6, 2021
 #1
avatar+112839 
+2

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 :)

 Mar 6, 2021
 #2
avatar
+2

The cosine rule?!?indecision

 

BD = sqrt(122 + 42 - 2*12*4 * cos(BCD))       and then what?!?wink

Guest Mar 6, 2021
edited by Guest  Mar 6, 2021
 #6
avatar+69 
+1

It's called the Law of Cosines,

OofPirate  Mar 7, 2021
 #3
avatar+112839 
+3

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}}\\
 

 Mar 7, 2021
 #4
avatar+1470 
+1

Excellent work, Melody!!!  smiley

Dragan  Mar 7, 2021
edited by Dragan  Mar 7, 2021
 #5
avatar+112839 
+1

Thanks Dragen   laugh

 

I really do think that there must be a better way to tackle this problem though.

Melody  Mar 7, 2021
 #7
avatar+1470 
+2

There's more than one way to... smiley

.................................................................

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.laugh 

Dragan  Mar 7, 2021
edited by Dragan  Mar 7, 2021
 #8
avatar+112839 
+1

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)

Melody  Mar 7, 2021
 #11
avatar+2197 
+1

aHeM frowncheeky

CalTheGreat  Mar 8, 2021
 #12
avatar+112839 
0

Hi Cal  :)

Melody  Mar 8, 2021
 #13
avatar+2197 
+2

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. 

CalTheGreat  Mar 8, 2021
 #9
avatar
+2

AB.CD + BC.AD = AC.BD

is known as Ptolemy's theorem,

dates back around two thousand years.

 Mar 7, 2021
 #10
avatar+112839 
+1

Thanks for the info guest.    It is always good to get interesting background information.

 

Jugoslav was still asked to prove it though. 

He couldn't just say. It is true because Ptolomy said so.

Melody  Mar 7, 2021
 #14
avatar+1164 
0

Hahahaha...

jugoslav  Mar 8, 2021
 #15
avatar+69 
0

Ptolemy is always correct: Because the Party and Big Brother says so

 

In the same logic: Every single math problem has an answer of 53

OofPirate  Mar 13, 2021

64 Online Users

avatar
avatar
avatar
avatar
avatar