Question

In the given figure, ABCD is a parallelogram. P is any point in its interior.

Prove that : ar (ΔAPD) + ar (ΔBPC) = ar (ΔAPB) + ar (ΔCPD)

Explain with complete calculations & justifications.

Points : 30 ☺

Answer 1

We know that,
             Area of a Δ is (1/2)*base*height.
So with reference to the figure,
      Ar(ΔAPD)=(1/2)*x*DP...................(1)
     Ar(ΔBPC)=(1/2)*y*BP..................(2)
Since height for ΔAPB is x and for ΔCPD is y
So   Ar(ΔAPB)=(1/2)*x*BP...................(3)
      Ar(ΔAPD)=(1/2)*y*DP..................(4)
Thus from (1),(2),(3) and (4),
Ar(ΔAPD)*Ar(ΔBPC)=Ar(ΔAPB)*Ar(ΔAPD)

Answer 2

CONSTRUCTION :- draw a st. line MN parallel to AB and CD and passing through P.
AND
draw a st. line XY parallel to AD and BC passing through P.

now we know
area of llgm = 2(area of triangle)
when both the llgm and triangle have same base and lie within same parallels

therefore
area of llgm ABNM
=2(area of triangle APB) ----------(1)

area of llgm MNCD
= 2(area of triangle PCD)------------(2)

area of llgm AXYD
= 2( area of triangle APD) --------------(3)

area of llgm BXYC
= 2( area of triangle BPC)-----------(4)

by adding (1) and (2)
area of llgm ABNM +area of llgm MNCD
=2(area of triangle APB)+2(area of triangle PCD)

=area of llgm ABCD
= 2(area of triangleAPB + area of trianglePCB)--------(5)

by adding (3) and (4)
area of llgm AXYD +area of llgm BXYC
= 2( area of triangle APD)+2( area of triangle BPC)

area of ABCD
=2(area of APD + area of BPC)-------(6)

by comparing (5) and (6) we get
2(area of triangleAPB + area of trianglePCB)
= 2(area of triangle APD + area of triangle BPC)

area of triangleAPB + area of trianglePCB
= area of APD + area of BPC