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Question

In the figure prove that A>CandB<D
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Solution



In ΔABC, we have
BC>AB
[ AB is the smallest side]
BAC>BCA   ...(1)
Similarly, in ΔACD, we have
CD>AD
[ CD is the largest side]
CAD>ACD     ...(2)
On adding (1) and (2), we get
BAC+CAD>BCA+ACDBAD>BCDA>C
Now, in ΔABC, we have
AD>AB
ABD>ADB    ...(3)
[  CD is the largest side]
DBC>BDC
On adding (3) and (4), 23 get
ABD+DBC>ADB+BDCABC>ADCB>D
Hence, it is proved that A>CandB<D

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