Given: △abc, ab = cb ak ⊥ bc , mc ⊥ ab prove: ak = mc

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 ΔSAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and   including angle in the 2nd ΔASA ⇒ 2 angles and the side whose joining them in the 1st Δ  ≅ 2 angles and the side whose joining them in the 2nd ΔAAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles   and one side in the 2nd ΔHL ⇒ 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

#LearnwithBrainly

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4x-10=2x+3 +10. +104x=2x+13-2x. -2x2x=13/2. /2x=6.5