diagonals ac and bd of a quadrilateral abcd intersect at p show that ar(apb)×ar(cpd)=ar(apd)×ar(bpc)
Answers
Answered by
14
Hello,
we have that,
area of triangle= 1/2×base×altitudine;
so:
ar(apb)×ar(cpd)=(1/2×bp×am)×(1/2×pd×cn)
= 1/4×bp×am×pd×cn
ar(apd)×ar(bpc)=(1/2×pd×am)×(1/2×cn×bp)
=1/4×pd×am×cn×bp
=1/4×bp×am×pd×cn
then
ar(apb)×ar(cpd)=ar(apd)×ar(bpc)
bye :-)
we have that,
area of triangle= 1/2×base×altitudine;
so:
ar(apb)×ar(cpd)=(1/2×bp×am)×(1/2×pd×cn)
= 1/4×bp×am×pd×cn
ar(apd)×ar(bpc)=(1/2×pd×am)×(1/2×cn×bp)
=1/4×pd×am×cn×bp
=1/4×bp×am×pd×cn
then
ar(apb)×ar(cpd)=ar(apd)×ar(bpc)
bye :-)
harshita73:
Thanks
Similar questions