Question

ABCD is quadrilateral in which AC bisect it into two equal areas. Prove that AC bisects BD

Answer 1

Draw BE ⊥ AC.   and   DF ⊥ AC.    
Ar(ΔABC) = Ar(ΔADC) 
1/2 * AC * BE = 1/2 * AC * DF     =>  BE = DF

Let AC and BD intersect at O.
compare ΔOBE and ΔODC.
∠BOE = ∠DFC = vertically opposite angles
∠BEO = ∠DFO = 90°

Hence, the two triangles are similar.  
As  BE = DF, both triangles are congruent.
So  BO = OD .  Hence AC bisects BD.