Question

diagonals AC and BD of a quadrilateral ABCD intersect each other at O prove that
➰AB + BC + CD + D A> AC + BD


➰AB + BC + CD + DA< 2(AC + BD)

Answer 1

hey



Given:ABCD is a Quadrilateral

To prove :1)AB+BC+CD+DA>AC+BD
2)AB+BC+CD+DA<2(AC+BD)


PROOF: IN QUADRILATERAL ABCD,

IN triangle ABC, AB+BC>AC. ..1
TRIANGLE BCD, BC+CD>BD. ..2
TRIANGLE CDA,CD+DA>AC. ..3
TRIANGLE. DAB,DA+AB>BD. ..4


ON ADDING,(1) AND(2),

2(AB+BC+CD+DA)>2(AC+BD)
THUS,

AB+BC+CD+DA>AC+BD


2)

IN TRIANGLE OAB,OA+OB>AB. ...1
IN TRIANGLE OBC,OB+OC>BC. ...2
IN TRIANGLE OCD,OC+OD>CD ....3
IN TRIANGLE ODA,OD+OA>DA. ...4

ON ADDING THESS EQUATIONS



2(AC+BD)>AB+BC+CD+DA
THUS, AB+BC+CD+DA<2(AC+BD)



THANKS

JAY HIND

Answer 2

Answer:

here's the answer

hope you understand my handwriting !!!