Answer:
Δs ABC and ADC are congruent using the ASA postulate of congruence
Step-by-step explanation:
Let us revise the cases of congruence
- SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ
- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
- AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles and one side in the 2nd Δ
- HL ⇒ hypotenuse and leg of the 1st right Δ ≅ hypotenuse and leg of the 2nd right Δ.
In the given figure
∵ AC bisects ∠BAD and ∠BCD
→ That means AC divides ∠BAD into 2 equal angles ∠BAC and ∠DAC, and
divides ∠BCD into 2 equal angles ∠BCA and ∠DCA
∴ m∠BAC = m∠DAC
∴ m∠BCA = m∠DCA
In Δs ABC and ADC
∵ m∠BAC = m∠DAC
∵ m∠BCA = m∠DCA
∵ AC is a common side in the two triangles
∵ AC joining the congruent angles
→ By using the 3rd rule above
∴ Δs ABC and ADC are congruent using the ASA postulate of congruence
