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AD is the median of ?ABC, O is any point on AD. BO and CO produced meet AC and AB in E and F respectively. AD is produced to X such that OD = DX. Prove that AO : AX = AF : AB.

Tutor

In the rectangle OBXC, OD=DX and AB=DC given. So it a parrallelogram. Hene FB is parallel to CX and OC(Hence FC is parallel to BX. Hence in the traingle ABX, AO:AX = AF: AB Proved

Math Magician

ABC is a triangle, AD is median. O is a point on AD. Produced AD to X, OD = DX given. BD = CD, join BX and CX. Thus diagonals of quadrilateral OBXC bisect each other. So OBXC is a parallelogram. So BX parallel to CF. So BX parallel to OF. Now in triangle ABX, AO/AX = AF/AB =AO:AX = AF:AB Proved! read more

ABC is a triangle, AD is median. O is a point on AD. Produced AD to X, OD = DX given.

BD = CD, join BX and CX. Thus diagonals of quadrilateral OBXC bisect each other.

So OBXC is a parallelogram. So BX parallel to CF. So BX parallel to OF.

Now in triangle ABX, AO/AX = AF/AB

=AO:AX = AF:AB Proved!

5:8

Square root 3 upon 2

Chemistry Teacher

Sin^2(A)+ cos^2(A)=1 Cos^2(A)=1-Sin^2(A) =1-(1/2)^2 =1-(1/4) =3/4 Cos(A). =?(3/4) Cos(A). =(?3)/2

Science Teacher

I answered the question in the following attachment.

Math Tutor

The theorem is given in the pic

Teacher

Sketch the triangle and mark the points on it. In the rectangle OBXC, OD=DX and AB=DC given. So it a parrallelogram. Hene FB is parallel to CX and OC(Hence FC is parallel to BX. Hence in the traingle ABX, AO:AX = AF: AB Proved

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