Question

In parallelogram ABCD of the accompanying diagram, line DP is drawn bisecting BC at N and meeting AB (extended) at P. From vertex C, line CQ is drawn bisecting side AD at M and meeting AB (extended) at Q. Lines DP and CQ meet at O. Show that the area of triangle QPO is `9/8` of the area of the parallelogram ABCD

Answer 1

Draw OX perpendicular to QP.

In ΔADP, MN = `1/2` AP,

In ΔBCQ, MN = `1/2` QB

So, AP = BQ or AB + BP = AB + QA

∴ PB = QA

∴ QA = AB = BP or QP = QA + AB + BP = 3AB

Area of ΔOQP = `1/2 xx "QP" xx "OX"`

= `1/2 xx 3"AB" xx "OX"`

= `3/2 xx "AB" xx "OX"` 

= `3/2 "AB" ("OY" + "YX")`

= `3/2 xx "AB" xx "OY"  + 3/2 xx "AB" xx "YX"   ...("AB" = "MN")`

= `3/2 xx "MN" xx "OY" + 3/2 xx "AB" xx "YX"`

= `3 "Area ΔOMN" + 3/2 + "Area ΔBNM"`

= `3[1/4 "area of MNCD"] + 3/2 [1/2 "area of ABCD"]`

= `3/4[1/2 "area of ABCD"] + 3/4["area of ABCD"]`

= `3/8 "area of ABCD" + 3/4 "area of ABCD"`

= area of ABCD `[3/8 + 3/4]`

= area of ABCD `((3 + 6)/8)`

= `9/8` area of ABCD.

Hence it is proved.