Pawan Prajapati
Last Activity: 4 Years ago
(i) Consider triangles ADC and ABC,
L DAC = L BAC
[AC is bisector of LA]
L DCA = L BCA [AC is bisector of LC] AC is common.
.. 4ADC MBC
AD = AB.
[ASA]
[CPCTJ
As in rectangle ABCD, adjacent sides are equal. Hence ABCD is a square.
(ii) Consider triangles DAB and BCD,
AB = BC = CD = DA [Sides of a square]
BD is common.
4 DAB 4 DCB
L ADB = L CDB
[SSS!
[CPCTJ
114
and L ABD = L CBD
:. BD bisects LB and LD.
[CPCT)
[Using above results]