Question

Prove this!

apt answer should be

brainliest answer will be selected!!!
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Answer 1

In a rhombus all sides are equal and the diagonals bisect each other at right angles.

AB = BC = CD = DA; And AC ⊥ BD.

So from the above,

if O is the intersecting point of the two diagonals AC & BD, then,

OA = (1/2)AC & OB = (1/2)BD;

as well OAB is a right triangle with <AOB = 90 deg.

By Pythagoras theorem

OA² + OB² = AB²

[(1/2)AC]² + [(1/2)BD]² = AB²

(1/4)AC² + (1/4)BD² = AB²

Multiplying by 4,

AC² + BD² = 4AB² [Proved]

Answer 2

ABCd is rombhus

Ab=bc=ac=ad

AC and bc are perpendicular to each other and ao= 2ac bo=2bc

in triangle Aob, angle Aob= 90 ab^2= ao^2+bo^2 phytagoras theorem

(ac/2)^2+(bc/2)^2=ab^2

ab^2=ac^2/4+bc^2+4

ab^2=ac^2+bc^2/4

4ab^2=ac^2+bc^2