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In equilateral ∆ABC, points M, P, and K belong to AB , BC , and AC respectively and AM:MB = BP:PC = CK:KA = 1:3. Prove that ∆MPK is an equilateral triangle.

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Answer:

See explanation

Step-by-step explanation:

In equilateral ∆ABC,

[tex]AB = BC = AC=a[/tex]

[tex]m\angle A=m\angle B=m\angle C=60^{\circ}[/tex]

Points M, P, and K belong to AB , BC , and AC respectively and

AM:MB = BP:PC = CK:KA = 1:3.

So,

[tex]AM = BP = CK =\dfrac{1}{4}a[/tex]

[tex]MB = PC = KA =\dfrac{3}{4}a[/tex]

Triangles AMK, BPM and CKP are all congruent by SAS postulate, so

[tex]MK=MP=PK[/tex]

If [tex]MK=MP=PK,[/tex] triangle MPK is equilateral triangle

Ver imagen frika

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