Question

ABCD is cyclic quadrilateral, lines AB and DC intersect
in the point F and lines AD and BC intersect in the point
E. Show that the circumcircles of triangleBCF and triangleCDE
intersect in a point G on the line EF.


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Answer 1

FGE is straight line .

Step-by-step explanation:

$\angle FBC=90^{o}(Angle\, in\, the\ semi\, circle\, is\, 90^{0})\\\\\angle FGC=90^{o}\\\\Therefore, \angle FBC+\angle FGC=90^{o}+90^{o}=180^{o}\\\\Hence, FBCG\, is\,\,a\,cyclic\,quadrilteral.\\\\Similarly, DEGC \,is\, a \,cyclic \,quadrilteral\\\\\angle FGC+\angle EGC$

$=180^{o}-\angle FBE+180^{o}-\angle EDF (FBGC\,and\, DEGC\,are\,cyclic\,quadrilateral)\\\\=\angle ABE+\angle ADF (Linear pair)\\\=180^{o} (ABCD \,is\, a\, cyclic\, quadrilateral)\\\Hence,\,FGE\,is,straight\,line.$

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