Question

37.
in triangle ABC,AD is a median and O is any point on AD. BO and Co on producing meet AC and AB at E and F respectively. Now AD is produced to X such that OD = DX .
Prove that: (a)EF|| BC (b) AC:AX=AF:AB​

Answer 1

Median = AD (Given)

Point on AD = O (Given)

BC and OX bisect each other (Given)

Thus, BXCO is a parallelogram and BE || XC and BX || CF

In ΔABX, by B.P.T, Theorum

AF/FB = AO/OX -- 1

In ΔAXC

AE/EC = AO/ OX -- 2

From equation 1 and 2

AF/FB = AE/EC

Therefore, by the converse of BPT -

EF || BC

Now, as OX/AO = FB/AF - from 1

Adding 1 on both sides we will get -

AX/OA = AB/AF

Thus,

AC:AX=AF:AB​

Answer 2

Answer:

Step-by-step explanation:

BC and OX bisect each other (Given)

Thus, BXCO is a parallelogram and BE || XC and BX || CF

In ΔABX, by B.P.T, Theorum

AF/FB = AO/OX -- 1

In ΔAXC

AE/EC = AO/ OX -- 2

From equation 1 and 2

AF/FB = AE/EC

Therefore, by the converse of BPT -

EF || BC

Now, as OX/AO = FB/AF - from 1

Adding 1 on both sides we will get -

AX/OA = AB/AF

Thus,

AC:AX=AF:AB​