Answer:
The diagonals of the rhombus ABCD bisect each other
Step-by-step explanation:
Given: ABCD is a rhombus and AC and BD are diagonals. All sides of the rhombus are equal.
To prove: AC and BD bisects each other.
Let O is the intersection point diagonals.
In triangle ABC, ABD, BCD and ADC are isosceles triangles.
In triangle AOB and COD,
[tex]\angle DBA=\angle BDC[/tex]
[tex]\angle COD=\angle AOB[/tex] (Vertically opposite angle)
[tex]AB=DC[/tex] (opposite side of rhombus)
By AAS postulate,
[tex]\triangle COD\cong \triangle AOB[/tex]
[tex]AO=OC[/tex] (CPCTC)
O is the midpoint of segment AC. Segment BD bisects segment AC.
In triangle AOD and COB,
[tex]\angle BDA=\angle DBC[/tex]
[tex]\angle COB=\angle AOD[/tex] (Vertically opposite angle)
[tex]AD=BC[/tex] (opposite side of rhombus)
By AAS postulate,
[tex]\triangle AOD\cong \triangle COB[/tex]
[tex]BO=OD[/tex] (CPCTC)
Segment BO is congruent to segment OD.
O is the midpoint of segment BD. Segment AC bisects segment BD.
Hence proved.
