Given
AB  and  CD  intersect
AC,  CB,  BD  and  AD  are congruent.
Prove that AB  is the bisector of ∠CAD and ray  CD  is the bisector of ∠ACB.
and AB  and  CD  are perpendicular.
To proof
Bisector
A bisector is that which cut an angle in two equal parts.
In ΔACB and ΔADB
AD = AC Â ( Given )
AB = AB Â ( common )
BC = DB Â ( Given )
by SSS congurence property
we have
ΔACB ≅ΔADB
∠CAB =∠DAB
∠CBA = ∠DBA
( By corresponding sides of the congurent triangle )
Thus AB is the bisector of the ∠CAD.
InΔ DAC and ΔDBC
AD = DB (Given)
AC = CB Â ( Given )
CD = CD (common)
By SSS congurence property
ΔDAC≅ Δ DBC
∠ ACD =∠BCD
∠ADC =∠BDC
( By corresponding sides of the congurent triangle )
Therefore CD is the bisector of the CAD.
In ΔBOC andΔ BOD
BO = BO ( Common )
∠BCO = ∠BDO
( As prove above ΔACB ≅ΔADB
Thus ∠ACB = ∠ADB by corresponding sides of the congurent triangle , CD is a bisector
∠BCO = ∠BDO )
 CB = DB ( given )
by SAS congurence property
ΔBOC ≅ ΔBOD
∠BOC =∠BOD
∠BOC +∠BOD = 180 °( Linear pair )
2∠BOC = 180°
∠BOC = 90°
∠BOC =∠BOD = 90°
also
In ΔCOA and ΔAOD
AO = AO ( Common )
∠ACO =∠ADO
(  As prove above ΔACB ≅ΔADB Thus ACB = ADB by corresponding sides of congurent triangle ,CD is a bisector
thus  ∠ACO = ∠ADO )
AC =AD ( given )
by SAS congurence property
Δ COA ≅ ΔAOD
∠AOC = ∠AOD
( By corresponding angle of corresponding sides )
∠AOC + ∠AOD = 180°
2∠AOC = 180°  ( Linear pair )
∠AOC = 90°
∠AOC = ∠AOD = 90 °
Thus AB  and  CD  are perpendicular.
Hence proved
 Â
Â