Question

THE line segment XY is parallel to side AC of triangle ABC and divides the triangle into two parts of equal areas.
find the ratio AX/AB

Answer 1

Answer:

(AX/AB) = ((√2 – 1)/ √2)

Step-by-step explanation:

Given, Δ ABC, XY || AC, and ar(Δ BXY) = ar(AXYV)

Now,

In Δ ABC and Δ XBY,

angle ABC = angle XBY (common)

angle ACB = angle XYB  (∵ XY || AC, corresponding angles are equal)

Δ ABC  ~  Δ XBY (AA similarity)

We know that, in similar triangles ratio of are of triangle = ratio of square of corresponding sides

∴ (Area of Δ ABC)/(Area of Δ XBY) = (AB/XB)^2

(Area of Δ XBY + Area of AXYC)/(Area of Δ XBY) = (AB/XB)^2

(Area of Δ XBY + Area of Δ XBY)/ (Area of Δ XBY) = (AB/XB)^2

2 * ((Area of Δ XBY)/ (Area of Δ XBY)) = (AB/XB)^2

2 = (AB/XB)^2

(AB/XB) = √2

XB = (AB/√2)

And, ∵ AX + XB = AB

AX + (AB/√2) = AB

AX = AB - (AB/√2) = AB(1-(1/√2))

∴ (AX/AB) = ((√2 – 1)/ √2)

Answer 2

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