Question

line-segment AB is parallel to another line-segment CD. O is mid-point of AD . show that 1 triangle AOB~= triangle DOC 2 o is also the mid point of BC​

Answer 1

Answer:

n triangle AOB & triangle DOC

Angle OAB = angle ODC [alt.int. angles]

Angle SOB =angle DOC[V.O.A]

AO=OD(given)

Triangle SOB congruent to triangle DOC(A.S.A)

OB=OC(c.p.c.t.c)

Hence o is the mid point of BC

THIS answer is correct

Step-by-step explanation:

Given: AB = CD ; O is the mid - point .

To proof : ABC is congruent to DOC

Proof : IN triangle AOB and DOC ,

AB = CD ( given )

<OAB = <ODC ( alternate interior angles )

<AOB = <DOC ( vertically opposite angles )

Therefore , triangle AOB is congruent to DOC (ASA rule )

OB = OC ( C.PC.CT)

Hence , O is also the mid point of BC

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