A quadrilateral `A B C D` is such that diagonal `B D` divides its area in two equal parts. Prove that `B D` bisects `A Cdot` GIVEN : A quadrilateral `A B C D` in which diagonal `B D` bisects it. i.e. `a r( A B D)=a r( B D C)` CONSTRUCTION : Join `A C` Suppose `A C` and `B D` intersect at `O` . Draw `A L_|_B D` and `C M B Ddot` TO PROVE : `A O=O Cdot`

Explanation

Given: A quadrilateral `ABCD` in which diagonal `BD` bisects it.
`=>ar(/_\ABD)=ar(/_\BDC)`
To prove: `BD` bisects `AC`.
Construction: Join `AC`. Suppose `AC` and `BD` intersect at `O`. Draw `AL` and `CM` perpendicular to `BD`.
Proof: `triangleABD` and `triangleABC` are on the same base `AB` and have equal area.
`therefore` Their corresponding altitudes are equal i.e. `AL=CM`
Now,
In `triangleALO` and `triangleCMO`,
`angle1=angle2` (vertically opposite angles)
`angleALO= angleCMO` (right angles
`AL=CM`
`therefore triangleALO cong triangleCMO` (By `AAS`)
`=>AO=OC`
`BD` bisects `AC`.