Question

Complete the proof.

Answer 1

Answer:

△ AMC ≅ △ BMC ⇒ proved down

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 Δ ABC

∵ CM ⊥ AB

∴ m∠AMC = m∠BMC = 90°

∴ ∠1 ≅ ∠2

In the two triangles AMC and BMC

∵ ∠1 ≅ ∠2 ⇒ proved

∵ ∠3 ≅ ∠4 ⇒ given

∵ CM is a common side in the two triangles

∴ CM ≅ CM ⇒ common side

→ Two angles and the side joining them in the 1st triangle ≅ two angles

   and the side joining them in the 2nd triangle, then use the 3rd case

   of congruency above

∴ Δ AMC ≅ Δ BMC ⇒ by using ASA postulate