Q9 of 77 Page 15

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:

ar(Δ APB) × ar(Δ CPD) = ar(Δ APD) × ar(Δ BPC).

Construction: Draw BQ perpendicular to AC

And,


DR perpendicular to AC


Proof: We have,


L.H.S = Area (ΔAPB) * Area (ΔCPD)


= (AP * BQ) * (PC * DR)


= ( * PC * BQ) * ( * AP * DR)


= Area (ΔBPC) * Area (ΔAPD)


= R.H.S


Therefore,


L.H.S = R.H.S


Hence, proved


More from this chapter

All 77 →
7

In Fig. 15.80, ABCD is a trapezium in which AB||DC. Prove that ar(Δ AOD) = ar(Δ BOC).

 8

In Fig. 15.81, ABCD and CDEF are parallelograms. Prove that

ar(Δ ADE) = ar(Δ BCF).


 10

In Fig. 15.82, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, Show that ar(Δ ABC) = ar(Δ ABD).

 11

If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.