In a trapezoid with bases of lengths a and b, a line parallel to the bases is drawn through the intersection point of the diagon
2 answers:
Answer:
Consider the trapezoid ABCD. In this trapezoid BC=a and AD=b.
Since triangles BOC and AOD are somilar, then
[tex] \dfrac{AO}{OC}=\dfrac{DO}{OB}=\dfrac{AD}{BC}=\dfrac{b}{a}. [\tex]
OC/AO = OB/DO = BC/AD=/ba
Triangles OAE and CAB are similar
EO/BC=b/(a+b)
EO=ab/(a+b)
again
OF=ba/(b+a)
and.
EF=2a/(a+b) is arequired length.
You might be interested in