Question

15. ABCD is a rhombus and its two diagonals meet at
O, show that
AD? + DC2+CB + BA2 = 4D0+4 AO?​

Answer 1

Answer:

Hence Proved

Step-by-step explanation:

LET AC AND BD INTERSECT AT A POINT O

IN TRIANGLE AOD

BY PYTHAGORAS

AB^2 = AO^2 + DO^2

ON MULTIPLYING BY 4 ON BOTH SIDES

4AB^2.=4AO^2 + 4DO^2

ALL SIDES OF RHOMBUS ARE EQUAL SO---

AB^2 + BC^2 + CD^2 + DA^2 = 4AO^2 +4 DO^2

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Answer 2

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