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Class 10 > Learn Mathematics > AD is the median of ?ABC, O is any point on AD. BO...
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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.

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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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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!
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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!

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5:8
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Square root 3 upon 2
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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
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Science Teacher

I answered the question in the following attachment.
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Math Tutor

The theorem is given in the pic
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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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