# ABC an equilateral triangle. D, E & F are points on AB, BC and AC such that AD = BE =CF = AB/3. BF, CD, AE intersect to form traingle PQR.Find the ratio of area of traingle PQR : area of...

ABC an equilateral triangle. D, E & F are points on AB, BC and AC such that AD = BE =CF = AB/3. BF, CD, AE intersect to form traingle PQR.

Find the ratio of area of traingle PQR : area of traingle ABC. (PQR is inside ABC.)

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ABC an equilateral triangle. D, E & F are points on AB, BC and AC such that AD = BE =CF = AB/3. BF, CD, AE intersect to form traingle PQR.

Find the ratio of area of traingle PQR : area of triangle ABC. (PQR is inside ABC.)

Let CD intersect AE at Q, CD intersect BF at P and AE intersect BF at R. Also, let the side length of triangle ABC be 3x.

Now triangles BRE and BCF are similar by AA similarity since they share an angle at B; for the other angle note the symmetry of the construction guarantees that angles BRE,CPF and AQD are congruent so angles Q,R, and P are congruent also and must be 60 degrees.

(1)Then BR/BC = RE/CF = BE/BF or BR/3x = RE/x = x/BF.

Consider triangle BFC. Drop the perpendicular from F to BC, naming the foot H. Then triangle FHC is a 30-60-90 triangle, so FH=(sqrt(3)/2)x. Using the Pythagorean Theorem on triangle BFH yields BF=sqrt(7)x. Thus the scale factor from triangle BRE to triangle BCF is 1/sqrt(7).

From (1) we get BR=3x/sqrt(7) and PF=RE=x/sqrt(7).

Therefore RP = sqrt(7)x-3x/sqrt(7) - x/sqrt(7) or RP=(3sqrt(7)x)/7.

Finally triangle QPR is similar to triangle ABC so the ratio of the areas is in the ratio of the square of the side lengths.

Area QPR/Area ABC = [(3xsqrt(7)/7)/3x]^2=1/7.

The ratio of the area of triangle QPR to the area of triangle ABC is 1:7.

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