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Question Number 137314 by liberty last updated on 01/Apr/21

$$ \\ $$ The diagonals of a convex quadrilateral divide the area into A, B, D and C (numbered clockwise) such that A/B = C/D. Does this theorem have a name?\\n

Commented bymr W last updated on 01/Apr/21

it′s so obviours that i think there is  no special name for it. there is also  no special name for the theorem that  the sum of all internal angles of a  quadrilateral is 360°.

$${it}'{s}\:{so}\:{obviours}\:{that}\:{i}\:{think}\:{there}\:{is} \\ $$ $${no}\:{special}\:{name}\:{for}\:{it}.\:{there}\:{is}\:{also} \\ $$ $${no}\:{special}\:{name}\:{for}\:{the}\:{theorem}\:{that} \\ $$ $${the}\:{sum}\:{of}\:{all}\:{internal}\:{angles}\:{of}\:{a} \\ $$ $${quadrilateral}\:{is}\:\mathrm{360}°. \\ $$

Answered by EDWIN88 last updated on 01/Apr/21

Sure it has a name , Dunkley′s Theorem.  ′ The diagonals of convex quadrilateral ABCD  divide the quadrilateral into four triangles  ; the products of areas of the opposing   triangles are equal.′

$$\mathrm{Sure}\:\mathrm{it}\:\mathrm{has}\:\mathrm{a}\:\mathrm{name}\:,\:\mathrm{\color{mathred}{D}\color{mathred}{u}\color{mathred}{n}\color{mathred}{k}\color{mathred}{l}\color{mathred}{e}\color{mathred}{y}}\color{mathred}{'}\mathrm{\color{mathred}{s}}\color{mathred}{\:}\mathrm{\color{mathred}{T}\color{mathred}{h}\color{mathred}{e}\color{mathred}{o}\color{mathred}{r}\color{mathred}{e}\color{mathred}{m}}\color{mathred}{.} \\ $$ $$\color{mathred}{'}\color{mathred}{\:}\mathrm{The}\:\mathrm{diagonals}\:\mathrm{of}\:\mathrm{convex}\:\mathrm{quadrilateral}\:\mathrm{ABCD} \\ $$ $$\mathrm{divide}\:\mathrm{the}\:\mathrm{quadrilateral}\:\mathrm{into}\:\mathrm{four}\:\mathrm{triangles} \\ $$ $$;\:\mathrm{the}\:\mathrm{products}\:\mathrm{of}\:\mathrm{areas}\:\mathrm{of}\:\mathrm{the}\:\mathrm{opposing}\: \\ $$ $$\mathrm{triangles}\:\mathrm{are}\:\mathrm{equal}.'\: \\ $$

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