D,E and F are the midpoints of BC,CA,AB of triangle ABC. Show that ar(BDEF) =½ar(ABC)
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in Δabc
f and e r the mid points of ab and ac
so, ef // bc and ef is half of bc by mid point theorem
∴ bd = ef
hence, bdef is a //gm (a quadrilateral with a pair of opp sides equal and //)
similarly, cdef and aedf are also //gms
ar(dbf) = ar(def)(diagonals of a //gm divides it into 2 triangles with equal area)
similarly, ar(fde) = ar(dce)
ar(fde) = ar(aef)
ar(def) +ar(bdf) +ar(fde) +ar(dce) = ar(abc)
4 ar(def) = ar(abc)
ar(def) = ¹/₄ ar(abc)
ar(def) = ¹/₄ ar(abc)
ar(bdf) = ¹/₄ ar(abc)
ar(def) +ar(bdf) = ¹/₄ ar(abc) + ¹/₄ ar(abc)
ar(bdef) = ¹/₂ ar(abc)
f and e r the mid points of ab and ac
so, ef // bc and ef is half of bc by mid point theorem
∴ bd = ef
hence, bdef is a //gm (a quadrilateral with a pair of opp sides equal and //)
similarly, cdef and aedf are also //gms
ar(dbf) = ar(def)(diagonals of a //gm divides it into 2 triangles with equal area)
similarly, ar(fde) = ar(dce)
ar(fde) = ar(aef)
ar(def) +ar(bdf) +ar(fde) +ar(dce) = ar(abc)
4 ar(def) = ar(abc)
ar(def) = ¹/₄ ar(abc)
ar(def) = ¹/₄ ar(abc)
ar(bdf) = ¹/₄ ar(abc)
ar(def) +ar(bdf) = ¹/₄ ar(abc) + ¹/₄ ar(abc)
ar(bdef) = ¹/₂ ar(abc)
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