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
<em>A bisector is that which cut an angle in two equal parts.</em>
In ΔACB and ΔADB
AD = AC ( Given )
AB = AB ( common )
BC = DB ( 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 ( 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
, where 
is a binomial distribution with 
and 
. So
