The diagonals AC and BD of a rhombus intersect each other at O. Prove that:AB2+BC2+CD2+DA2+=4 (OA2+OB2)
Answers
Diagonal's of Rhombus bisect at 90° and AO=OC & BO=OD
Apply pythogorous
AB²=AO²+BO²
BC²=BO²+OC²
CD²=OD²+OC²
AD²=OD²+AO²
AB²+BC²+CD²+AD²= [2AO²+2BO²+2CO²+2DO²]
AB²+BC²+CD²+AD²= 2[(AO+OC)²-2AO.OC+(BO+OD)²-2BO.OD]
AB²+BC²+CD²+AD²=2[(4OA²)-2OA²+(4OB²)-2OB²]
AB²+BC²+CD²+AD²=2[2OA²+2OB²]
AB²+BC²+CD²+AD²=2×2[OA²+OB²]
AB²+BC²+CD²+AD²=4[OA²+OB²]
AB² + BC² + CD² + DA² = 4(OA² + OB²)
Step-by-step explanation:
The diagonals AC and BD of a rhombus intersect each other at O.
Diagonal of rhombus bisect each other perpendicularly
=> OA = OC = AC/2
& OB = OD = BD/2
Applying Pythagorean theorem
AB² = OA² + OB²
BC² = OB² + OC²
CD² = OC² + OD²
DA² = OD² + OA²
Adding all
=> AB² + BC² + CD² + DA² = OA² + OB² + OB² + OC² + OC² + OD² + OD² + OA²
using OA = OC & OB = OD
=>AB² + BC² + CD² + DA² = OA² + OB² + OB² + OA² + OA² + OB² + OB² + OA²
=> AB² + BC² + CD² + DA² = 4OA² + 4OB²
=> AB² + BC² + CD² + DA² = 4(OA² + OB²)
QED
Proved
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