By using congruence of two triangles AKC and CMA we proved that
AK = MC
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 1st Δ ≅ 2 angles
and one side in the 2nd Δ
- HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ
In Δ ABC
∵ AB = BC
∴ Δ ABC is an isosceles triangle
∴ ∠BAC ≅ m∠BCA ⇒ base angles of isoscelesΔ
∵ AK ⊥ BC
∴ m∠AKC = 90°
∵ MC ⊥ AB
∴ m∠CMA = 90°
∴ ∠AKC ≅ ∠CMA
In the two Δs AKC and CMA
∵ ∠BAC ≅ m∠BCA
∵ ∠AKC ≅ ∠CMA
∵ AC ≅ CA
∴ Δ AKC and Δ CMA are congruent by AAS postulate
∴ AK ≅ CM
By using congruence of two triangles AKC and CMA we proved that
AK = MC
Learn more:
You can learn more about congruence in brainly.com/question/3202836
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