Question

In a trapezoid with bases of lengths a and b, a line parallel to the bases is drawn through the intersection point of the diagonals. Find the length of the segment that is cut from that line by the legs of the trapezoid.

Answer 1

Answer:

$\dfrac{2ab}{a+b}.$

Step-by-step explanation:

Consider the trapezoid ABCD. In this trapezoid BC=a and AD=b.

Since triangles BOC and AOD are somilar, then

$\dfrac{AO}{OC}=\dfrac{DO}{OB}=\dfrac{AD}{BC}=\dfrac{b}{a}.$

Triangles OAE and CAB are similar, then

$\dfrac{EO}{BC}=\dfar{AO}{AC}=\dfrac{b}{a+b}.$

This means that

$EO=\dfrac{ba}{b+a}.$

Similarly, from similar triangles FDO and CDB:

$OF=\dfrac{ba}{b+a}.$

Thus,

$EF=\dfrac{2ab}{a+b}.$