Sum of the product of sides of cyclic quadrilateral is equal to product of diagonals
Answers
Answer:
Step-by-step explanation:
Converse. In a quadrilateral, if the product of its diagonals is equal to the sum of the products of the pairs of the opposite sides, then the quadrilateral is indescribable.
When the intersection is internal, the equality states that the product of the segment lengths into which P divides one diagonal equals that of the other diagonal. This is known as the intersecting chords theorem since the diagonals of the cyclic quadrilateral are chords of the circumcise.
Answer:
When the intersection is internal, the equality states that the product of the segment lengths into which P divides one diagonal equals that of the other diagonal. This is known as the intersecting chords theorem since the diagonals of the cyclic quadrilateral are chords of the circumcircle.
Step-by-step explanation:
This is the ptolemy theorem
Let ABCD be a random quadrilateral inscribed in a circle.
The proposition will be proved if AC\cdot BD = AB\cdot CD + AD\cdot BC.AC⋅BD=AB⋅CD+AD⋅BC.
It's easy to see in the inscribed angles that \angle ABD = \angle ACD, \angle BDA= \angle BCA,∠ABD=∠ACD,∠BDA=∠BCA, and \angle BAC = \angle BDC.∠BAC=∠BDC.
Let EE be a point on ACAC such that \angle EBC = \angle ABD = \angle ACD,∠EBC=∠ABD=∠ACD, then since \angle EBC = \angle ABD∠EBC=∠ABD and \angle BCA= \angle BDA,∠BCA=∠BDA,
\triangle EBC \approx \triangle ABD \Longleftrightarrow \dfrac{CB}{DB} = \dfrac{CE}{AD} \Longleftrightarrow AD\cdot CB = DB\cdot CE. \qquad (1)
△EBC≈△ABD⟺
DB
CB
=
AD
CE
⟺AD⋅CB=DB⋅CE.(1)
Note that \angle ABD = \angle EBC \Longleftrightarrow \angle ABD + \angle KBE = \angle EBC + \angle KBE \Rightarrow \angle ABE = \angle CBK.∠ABD=∠EBC⟺∠ABD+∠KBE=∠EBC+∠KBE⇒∠ABE=∠CBK. Then since \angle ABE= \angle CBK∠ABE=∠CBK and \angle CAB= \angle CDB,∠CAB=∠CDB,
\triangle ABE \approx \triangle BDC \Longleftrightarrow \dfrac{AB}{DB} = \dfrac{AE}{CD} \Longleftrightarrow CD\cdot AB = DB\cdot AE. \qquad (2)
△ABE≈△BDC⟺
DB
AB
=
CD
AE
⟺CD⋅AB=DB⋅AE.(2)
Therefore, from (1)(1) and (2),(2), we have
AB\CD + AD\BC = CE\DB + AE\DB = (CE+AE)DB = CA\DB.
AB⋅CD+AD⋅BC
=CE⋅DB+AE⋅DB
=(CE+AE)DB
=CA⋅DB.
Hence proved.