How do you prove Ptolemy's Theorem that in a cyclic quadrilateral, sum of the products of opposite pair of sides is equal to the product of the diagonals?

In cyclic quadrilateral #ABCD#, #ABxxCD+BCxxAD=ACxxBD#.

1 Answer
Shwetank Mauria
Feb 13, 2017

Please see below.

Explanation:

Let #ABCD# be the cyclic quadrilateral inscribed in a circle as decsribed below. Let us also join its diagonals #AC# and #BD# to intersect at #P#.
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To prove that #ABxxCD+BCxxAD=ACxxBD#, let us draw an angle #/_ABQ=/_CBD# to intersect #AC# at #Q#, as shown.

Now in #DeltaABQ# and #DeltaCBD#, we have

#/_ABQ=/_CBD# - by construction
#/_BAQ=/_CDB# - angles in the same arc #BC#

#:.DeltaABQ~DeltaCBD#

and hence corresponding sides are proportional

#(AB)/(BD)=(AQ)/(CD)# or #ABxxCD=BDxxAQ# ..........(1)

Similarly as #/_ABQ=/_CBD#, we have #/_ABQ+/_QBP=/_CBD+/_QBP#

or #/_ABD=/_QBC# and in #DeltaABD# and #DeltaQBC#, we have

#/_ABD=/_QBC#
#/_ADB=/_QCB# - angles in the same arc #AB#

#:.DeltaABD~DeltaQBC#

and hence corresponding sides are proportional

#(AD)/(QC)=(BD)/(BC)# or #ADxxBC=BDxxQC# ..........(2)

Adding 1 and 2, we get

#ABxxCD+ADXXBC=BDxx(AQ+QC)=BDxxAC#

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