In the given figure- B C- then-AD - BCAD BCAD - 2BC

Question

In the given figure- B  C- then-AD - BCAD  BCAD - 2BC

Toppr Admin · Accepted Answer

In -x25B3-AOB- -x2220-A-gt-x2220-BHence- OB-gt-OA- -i-In -x25B3-ODC- -x2220-D-gt-x2220-CHence- OC-gt-OD- -ii-Adding inequalities -i- and -ii- we getOB-OC-gt-OA-ODBC-gt-AD-xA0-

In the given figure, $B < A$ and $D > C$ , then:

$A D = B C$
$A D < B C$
$A D > B C$
$A D = 2 B C$

In $△ A O B$ , $∠ A > ∠ B$
Hence, $O B > O A$ ..... (i)
In $△ O D C$ , $∠ D > ∠ C$
Hence, $O C > O D$ .... (ii)
Adding inequalities (i) and (ii), we get
$O B + O C > O A + O D$
$B C > A D$

In the given figure, B<A and D>C, then:AD=BCAD<BCAD>BCAD=2BC

In △AOB, ∠A>∠BHence, OB>OA..... (i)In △ODC, ∠D>∠CHence, OC>OD.... (ii)Adding inequalities (i) and (ii), we getOB+OC>OA+ODBC>AD

In the given figure, $B < A$ and $D > C$ , then:

$A D = B C$
$A D < B C$
$A D > B C$
$A D = 2 B C$

In $△ A O B$ , $∠ A > ∠ B$
Hence, $O B > O A$ ..... (i)
In $△ O D C$ , $∠ D > ∠ C$
Hence, $O C > O D$ .... (ii)
Adding inequalities (i) and (ii), we get
$O B + O C > O A + O D$
$B C > A D$