in a rhombus ABCD prove that AC square + BD square is equals to 4 every square
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Hey mate here is your answer
In ΔBOC,BC^2=BO^2+OC^2
Since BO=1/2BD,andOC=1/2AC
⇒BC^2=(1/2BD)^2+(1/2AC)^2
⇒BC^2=1/4(BD)^2+1/4(AC)^2
⇒BC^2=1/4(BD^2+AC^2)
⇒4BC^2=AC^2+BD^2
hope it helps you
Please mark as BRAINLIEST.
In ΔBOC,BC^2=BO^2+OC^2
Since BO=1/2BD,andOC=1/2AC
⇒BC^2=(1/2BD)^2+(1/2AC)^2
⇒BC^2=1/4(BD)^2+1/4(AC)^2
⇒BC^2=1/4(BD^2+AC^2)
⇒4BC^2=AC^2+BD^2
hope it helps you
Please mark as BRAINLIEST.
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