in the adjacent ligure, AB = CD. AB 1 BC
and DC ICBBC is bisected by AD at o
I AO = OOP Give reasons
Answers
Step-by-step explanation:
10th
Maths
Triangles
Basic Proportionality Theorem (Thales Theorem)
The side BC of a triangle A...
MATHS
The side BC of a triangle ABC is bisected at D; O is any point in AD. BO and CO produced meet AC and AB in E and F respectively and AD is produced to X so that D is the mid-point of OX. Prove that AO:AX=AF:AB and show that FE||BC.
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ANSWER
According to the question,
BD=CD and OD=DX
Therefore, BC and OX bisect each other and OBXC is a parallelogram.
This gives BX∥CO and CX∥BO
or BX∥CF and CX∥BE
or BX∥OF and CX∥OE
In ΔABX, as BX∥OF, then,
AX
AO
=
AB
AF
.............(1)
In ΔACX, as CX∥OE, then,
AX
AO
=
AC
AE
............. (2)
From equation (1) and (2),
AB
AF
=
AC
AE
Hence, as E and F divides AB and AC respectively in the same ratio, so, FE∥BC.