Segments ab and cd intersect (see picture to the right). prove that if segments ac , cb , bd and ad are congruent, then ray ab is the bisector of ∠cad and ray cd is the bisector of ∠acb. prove that segments ab and cd are perpendicular.

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  ( 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

(x+3)(x+7)

step-by-step explanation: