Finding diagonal $p$ of a cyclic quadrilateral with sides $(a,b,c,d)$

Question

Let the semi-perimeter of each divided triangle be
$$(2s_1,2s_2)= (a+b+p, c+d+p);$$
How to find the common diagonal $p?$ Equating common circum-diameter fetches
$$ dfrac{a^2b^2}{c^2d^2}=dfrac{s_1(s_1-a)(s_1-b)(s_1-p)}{s_2(s_2-c)(s_2-d)(s_2-p)}$$
s1 = (a + b + p)/2; s2 = (c + d + p)/2;
Solve[(a b/p/q)^2 == 
  s1 (s1 - a) (s1 - b) (s1 - p)/(s2 (s2 - c) (s2 - d) (s2 - p)), p]

As the expression has cyclic symmetry there may be easier ways to Simplify.
EDIT1:
After finding $q$ as well can we verify the (2000 years old) Ptolemy's relation?
$$;pq=ac+bd ;$$
Thanks for help.

Akku14 · Answer

Give support to Reduce by eliminating quotients since they are unequal zero.

f = Subtract @@ ((a^2 b^2)/(
  c^2 d^2) == ((s1 - a) (s1 - b) (s1 - p))/((s2 - c) (s2 - 
     d) (s2 - p))) // Together // Numerator

red = Reduce[{f == 0, 2 s1 == a + b + p, 2 s2 == c + d + p}, 
                p, {s1, s2}]

Simplify[red, {a > 0, b > 0, c > 0, d > 0}]

Or

Simplify[red, {0 < a < b < c < d}]

(*   p == Root[
 a^2 b^2 c^3 - a^2 b^2 c^2 d - a^2 b^2 c d^2 - a^3 c^2 d^2 + 
 a^2 b c^2 d^2 + a b^2 c^2 d^2 - b^3 c^2 d^2 + 
 a^2 b^2 d^3 + (-a^2 b^2 c^2 + 2 a^2 b^2 c d - a^2 b^2 d^2 + 
    a^2 c^2 d^2 - 2 a b c^2 d^2 + b^2 c^2 d^2) #1 + (-a^2 b^2 c - 
    a^2 b^2 d + a c^2 d^2 + b c^2 d^2) #1^2 + (a^2 b^2 - 
    c^2 d^2) #1^3 &, 1] || 
 p == Root[
 a^2 b^2 c^3 - a^2 b^2 c^2 d - a^2 b^2 c d^2 - a^3 c^2 d^2 + 
 a^2 b c^2 d^2 + a b^2 c^2 d^2 - b^3 c^2 d^2 + 
 a^2 b^2 d^3 + (-a^2 b^2 c^2 + 2 a^2 b^2 c d - a^2 b^2 d^2 + 
    a^2 c^2 d^2 - 2 a b c^2 d^2 + b^2 c^2 d^2) #1 + (-a^2 b^2 c - 
    a^2 b^2 d + a c^2 d^2 + b c^2 d^2) #1^2 + (a^2 b^2 - 
    c^2 d^2) #1^3 &, 2] || 
 p == Root[
 a^2 b^2 c^3 - a^2 b^2 c^2 d - a^2 b^2 c d^2 - a^3 c^2 d^2 + 
 a^2 b c^2 d^2 + a b^2 c^2 d^2 - b^3 c^2 d^2 + 
 a^2 b^2 d^3 + (-a^2 b^2 c^2 + 2 a^2 b^2 c d - a^2 b^2 d^2 + 
    a^2 c^2 d^2 - 2 a b c^2 d^2 + b^2 c^2 d^2) #1 + (-a^2 b^2 c - 
    a^2 b^2 d + a c^2 d^2 + b c^2 d^2) #1^2 + (a^2 b^2 - 
    c^2 d^2) #1^3 &, 3]   *)

Finding diagonal $p$ of a cyclic quadrilateral with sides $(a,b,c,d)$

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