Question

The diagonal AC and BD of a quadrilateral ABCD intersect at O and decide the quadrilateral ABCD into four triangle of equal area. Show that ABCD is a parallelogram

Answer 1

Step-by-step explanation:

Given: A quadrilateral ABCD where OB = OD & AB = CD

To prove: ar (DOC) = ar (AOB)

Proof: Let us draw DPI AC and BQ I AC.

In ADOPT and ΔΒΟ ,

ZDPO = ZBQO

ZIP = BOQ

OD = OB

ADOPT = A BOW

: DP = BQ

& ar(DOP) = ar(BOQ) =

(Both 90°)

(Vertically opposite angles)

(Given)

(AAS congruence rule)

(CPCT) (1)

(Area of congruent triangles is equal) .(2)

In ACDP and AABQ,

ZCPD = ZAQB

CD = AB

DP = BQ

- ACDP = AABQ

→ ar(CDP) = ar(ABQ)

Adding (2) & (3)

(Both 90°)

(Given)

(From (1)

(RHS congruence rule)

(Area of congruent triangles is equal) .(3)

ar(DOP) + ar(CDP) = ar(BOQ) + ar(ABQ)

ar (DOC) = ar (AOB)

Hence proved