Question

A line that goes through the point of intersection of the diagonals of a trapezoid divides one of the bases into two segments. The ratio of the lengths of these segments is m:n. What is the ratio of the lengths of the segments of the other base considering the segments that start with the same legs as the segments in the ratio for the first base?.

Answer 1

The ratio of the lengths of the segments of the other base considering the segments that start with the same legs as the segments in the ratio for the first base is n:m

What is ratio of two quantities?

Suppose that we've got two quantities with measurements as 'a' and 'b'

Then, their ratio(ratio of a to b) a:b or $\dfrac{a}{b}$

We usually cancel out the common factors from both the numerator and the denominator of the fraction we obtained. Numerator is the upper quantity in the fraction and denominator is the lower quantity in the fraction).

Suppose that we've got a = 6, and b= 4, then:

$a:b = 6:2 = \dfrac{6}{2} = \dfrac{2 \times 3}{2 \times 1} = \dfrac{3}{1} = 3\\or\\a : b = 3 : 1 = 3/1 = 3$

Remember that the ratio should always be taken of quantities with same unit of measurement. Also, ratio is a unit-less(no units) quantity.

What is Angle-Angle similarity for two triangles?

Two triangles are similar if two corresponding angles of them are of same measure. It is because when two pairs of angles are similar, then as the third angle is fixed if two angles are fixed, thus, third angle pair also gets proved to be of same measure. This makes all three angles same and thus, those two triangles are scaled copies of each other. Thus, they're called similar.

For the given case, referring to the attached figure, we can see that the lines AB and CD are parallel, the point of intersection of the diagonals is O, the line intersecting the bases is EF, the base whose segments' lengths' ratio are given is AB, and the base whose segments' lengths' ratio is to be found is CD.

We get $m\angle DOF = m\angle BOM$ vertical angles(angle made oppositely between two intersecting lines. They are always of same measure).

('m' of an angle means measurement of that angle. )

Also, the angles $m\angle OFD = m\angle OMB$ due to them being alternate interior angles(angles made by a line which intersects two parallel lines and those angles are opposite to each other on each parallel side, interior to those parallel sides. They are of same measure).

Thus, by Angle-Angle similarity, we get $\triangle BOM \sim \triangle DOF$

In the similar way, we get $\triangle AOM\sim \triangle COF$

Now, let for the similar triangle pair $\triangle BOM \sim \triangle DOF$, we have sides of BOM multiplied by 'a' to get sides of DOF.

And let for the similar triangle pair $\triangle AOM\sim \triangle COF$, we have sides of AOM multiplied by 'b' to get sides of COF.

Then, for the side OF and OE, we get:

$OF = a \times OM\\OF = b\times OM$

That shows that $a = b$

Thus, as we're given that:

$\dfrac{AE}{EB} = \dfrac{m}{n}$

So, we get:

$\dfrac{DF}{FC} = \dfrac{a \times BE}{a \times EA} = \dfrac{BE}{EA} = \dfrac{n}{m}\\\\\dfrac{DF}{FC} = \dfrac{n}{m}$

Thus, the ratio of the lengths of the segments of the other base considering the segments that start with the same legs as the segments in the ratio for the first base is n:m

Learn more about similar triangles here:

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