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.
enter image source here
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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