ABCDis a quadrilateral P and Q are the mid points of sides CD and AB respectivel AP and DQ - Maths - Areas of Parallelograms and Triangles

Question

ABCDis a quadrilateral P and Q are the mid points of sides CD
and AB respectivel AP and DQ meet at X whereas Bp and CQ meet at Y
Prove that area of ADX+area of BCY=area of quadrilateral PXQY

Ajanta Trivedi · Accepted Answer

<img alt="" src=
"https://img-nm%20mnimgs%20com/img/study_content/content_ck_images/images/pic81%20png"
style="width: 345px; height: 216px;">
given: P and Q are the mid-points of CD and AB respectively
TPT: area(ADX)+area(BCY)= area of quadrilateral PXQY
proof:
area (ADQ) = area (BDQ) (1) [since median divides a triangle into
two equal areas]
area(ADQ)-area(AQX) = area(BDQ)-area(AQX)
area(ADX) = area(BDQ) - area(AQX) (2)
similarly
area(BPC) - area(PYC) = area(BPD) - area(PYC)
area(BYC) = area(BPD) - area(PYC) (3)
adding eq(2) and eq(3):
area(ADX)+area(BYC) = area(BDQ)+area(BPD) -
[area(AQX)+area(PYC)]
= area(BPDQ) - [area(AQX)+area(PYC) (4)
area(DQP) = area(QPC) [since median divides triangle into equal
areas]
similarly :area(PQB) = area(APQ)
area(DQP) + area(PQB) = area(QPC) + area(APQ)
area(QBPD) = area(AQCP)
area(QBY)+area(DXP) = area(AQX)+area(PYC) (5)
from eq(4) and eq(5):
area(ADX)+area(BYC) = area(BPDQ) - area(QBY) - area(DXP)
= area of quadrilateral PXQY
hope this helps you

ABCDis a quadrilateral.P and Q are the mid points of sides CD and AB respectivel. AP and DQ meet at X whereas Bp and CQ meet at Y. Prove that area of ADX+area of BCY=area of quadrilateral PXQY.