Answer:
The statement If ∠A ≅ ∠C not prove that Δ ABD ≅ Δ CBD by SAS ⇒ C
Step-by-step explanation:
* Lets 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 first triangle ≅ 2 angles
and one side in the 2ndΔ
- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse
leg of the 2nd right angle Δ
* Lets solve the problem
- In the 2 triangles ABD , CBD
∵ AB = CB
∵ BD is a common side in the two triangles
- If AD = CD
∴ Δ ABD ≅ Δ CBD ⇒ SSS
- If BD bisects ∠ABC
∴ m∠ABD = m∠CBD
∴ Δ ABD ≅ Δ CBD ⇒ SAS
- If ∠A = ∠C
∴ Δ ABD not congruent to Δ CBD by SAS because ∠A and ∠C
not included between the congruent sides
* The statement If ∠A ≅ ∠C not prove that Δ ABD ≅ Δ CBD by SAS
