ABCD is a parallelogram. E is a point on AB such that 234×AE=EB. Let DE intersect AC at F. What is the ratio AC:AF?
Given:
Let r=AE:EB
Draw a diagram of the parallelogram ABCD.
Insert E on AB.
Extend DE to join CA at F.
(The above is as given by the question).
Examine the triangles FAE and DBE.
Show that they are similar noting that FA is parallel to DB.
Therefore the ration FE:ED = AE:EB
Similarly, show that triangles FAE and FCD are similar, noting again that AE is parallel to CD.
Hence show that
FE:AC = FE:ED
or equivalently,
AC:AF = ED:FD.
23
To find the ratio AC:AF, we need to determine the relationship between the lengths of the two segments.
Given that ABCD is a parallelogram, we know that opposite sides are parallel and congruent. Hence, AD || BC and AB || CD.
From the given information, we know that 234 × AE = EB. Let's represent AE as 'x'. Then, EB would be equal to 234x.
Since AD and BC are parallel, we can determine that triangles ADE and CBE are similar by the AA (Angle-Angle) postulate. This means that the corresponding sides of the two triangles are proportional.
Let's express the lengths of the corresponding sides:
AD / CB = AE / BE
AC + CD / CB = AE / BE
(Note: AC + CD is equal to AB by the definition of a parallelogram.)
AB / CB = AE / BE
Substituting the known values:
(AC + CD) / CB = x / 234x
Simplifying:
(AC + CD) / CB = 1 / 234
Cross-multiplying:
234(AC + CD) = CB
Now, we will use the fact that DE intersects AC at F:
By the segment theorem, we know that:
BC / CB = AF / FC
Since BC represents CD in this case, we substitute CD with BC:
BC / BC = AF / FC
1 = AF / FC
Thus, we have found that AF = FC.
Now, we can substitute the expression for BC from earlier:
234(AC + BC) = BC
234AC + 234BC = BC
233BC = 234AC
BC / AC = 234 / 233
Since AF = FC, we can say that:
AC / AF = 234 / 233
Therefore, the ratio AC:AF is 234:233.