P is a point in the interior of a parallelogram ABCD. Show that

(i) ar(∆APB) + ar(∆PCD) ar(ABCD)

(ii) ar(∆APD) + ar(∆PBC) = ar(∆APB) + ar(∆PCD)

(Hint : Through P, draw a line parallel to AB)

i) Construct segment GH parallel to AB and CD passing through point P as shown

Also construct perpendiculars PI and PJ on segments CD and AB respectively

Consider parallelogram DCGH

Base = CD …from figure

Height = PI …from figure

Area of parallelogram = base × height

Area of parallelogram DCGH = CD × PI …(i)

Consider parallelogram ABGH

Base = AB …from figure

Height = PJ …from figure

Area of parallelogram = base × height

Area of parallelogram ABGH = AB × PJ …(ii)

Area of triangle = × base × height

For ΔPCD

Base = CD

Height = PI

Area of ΔPCD = × CD × PI

Using (i)

⇒ Area of ΔPCD = × Area of parallelogram DCGH …(iii)

For ΔAPB

Base = AB

Height = PJ

Area of ΔAPB = × AP × PJ

Using (ii)

⇒ Area of ΔAPB = × Area of parallelogram ABGH …(iv)

Add (iii) and (iv)

⇒ Area of ΔPCD + Area of ΔAPB = × Area of parallelogram DCGH + × Area of parallelogram ABGH

⇒ Area of ΔPCD + Area of ΔAPB = × (Area of parallelogram DCGH + Area of parallelogram ABGH)

From figure Area of parallelogram DCGH + Area of parallelogram ABGH = area of parallelogram ABCD

Therefore, Area of ΔPCD + Area of ΔAPB = × area of parallelogram ABCD

⇒ area of parallelogram ABCD = 2 × Area of ΔPCD + 2 × Area of ΔAPB …(*)

ii) from figure

area(ΔDPC) = area(DCGH) – area(ΔDHP) – area(CPG) …(i)

area(ΔAPB) = area(ABGH) – area(ΔAPH) – area(BPG) …(ii)

add equation (i) and (ii)

⇒ area(ΔDPC) + area(ΔAPB) = [area(DCGH) + area(ABGH)] – [area(ΔDHP) + area(ΔAPH)] – [area(CPG) + area(BPG)]

⇒ area(ΔDPC) + area(ΔAPB) = area(ABCD) – area(APD) – area(BPC)

Using equation (*) from first part of question

⇒ area(ΔDPC) + area(ΔAPB) = 2 × Area of ΔPCD + 2 × Area of ΔAPB – area(APD) – area(BPC)

Rearranging the terms we get

area(APD) + area(BPC) = 2 × Area of ΔPCD + 2 × Area of ΔAPB - area(ΔDPC) - area(ΔAPB)

therefore, area(APD) + area(BPC) = area(ΔPCD) + area(ΔAPB)