posted by anoynomous .
in triangle ABC P, Q AND R ARE the mid points of side AB, BC, CA.prove that
area of parallelogram APQR=( 1/2)* area of triangle[ABC]
By the mid-point of triangle theorem,
PR is || to BC and PR = (1/2)BC
Also triange APR is similar to triange ABC
so the areas are proportional to the square of their sides.
since the sides are 1:2
their areas are 1:4
so triangle APR = (1/4) of triangle ABC
Similary BQP would be 1/4 of triangle ABC, and
RQC is 1/4 of triangle ABC, leaving the inside triangle PQR also as 1/4 of triangle ABC to get 4/4
so figure APQR = 2/4 or 1/2 of triangle ABC
How traingle ABC and APR are similar
Since BC and PR are similar
angle B = angle APR
and angle A is common
if 2 angles of a triangle are equal to 2 corresponding angles of another triangle, the triangles are similar.
Same argument for the other pairs of similar triangles.