Question -
Answer -
Given, ABCDis a rhombus whose diagonals AC and BD intersect at O.
We have to prove, as per the question,
AB2 + BC2 + CD2 +AD2 = AC2 + BD2
Since, the diagonals of a rhombus bisect eachother at right angles.
Therefore, AO = CO and BO = DO
In ΔAOB,
∠AOB = 90°
AB2 = AO2 + BO2 …………………….. (i) [By Pythagoras theorem]
Similarly,
AD2 = AO2 + DO2 …………………….. (ii)
DC2 = DO2 + CO2 …………………….. (iii)
BC2 = CO2 + BO2 …………………….. (iv)
Adding equations (i) + (ii) + (iii) + (iv), we get,
AB2 + AD2 + DC2 + BC2 =2(AO2 + BO2 + DO2 + CO2)
= 4AO2 + 4BO2 [Since, AO =CO and BO =DO]
= (2AO)2 +(2BO)2 = AC2 + BD2
AB2 + AD2 + DC2 + BC2 =AC2 + BD2
Hence, proved.