Given: ABCD is a rhombus
To prove: DB bisects ∠ABC and ∠ADC (∠1=∠2 and ∠3=∠4)
Proof: In ΔADB and ΔCDB
AD=CD Â Â Â Â Â Â Â (sides of same rhombus)
DB=DB Â Â Â Â Â Â Â (common in both triangle)
AB=CB Â Â Â Â Â Â Â Â (sides of same rhombus)
∴ ΔADB ≅ ΔCDB by SSS congruence property.
Angle bisector: A line divide an angle into two equal part.
CPCT: Congruent part of congruence triangles.
Match the statements:-
- ABCD  is rhombus  ⇒   Given
- ΔADB≅ΔCDB  ⇒ Diagonals of parallelogram make congruent Δ   Â
- ∠1=∠2,∠3=∠4  ⇒ CPCTE                 Â
- DB bisects ∠ABC and ∠ADC ⇒ Definition of angle bisector
Thus, Above matching is correct.