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SCORPION-xisa [38]
3 years ago
10

Mujer culona

Mathematics
1 answer:
KIM [24]3 years ago
6 0
Mujer culona mujer culona mujer culona
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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
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Answer:

See explanation

Step-by-step explanation:

In equilateral ∆ABC,

AB = BC = AC=a

m\angle A=m\angle B=m\angle C=60^{\circ}

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

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

So,

AM = BP = CK =\dfrac{1}{4}a

MB = PC = KA =\dfrac{3}{4}a

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

MK=MP=PK

If MK=MP=PK, triangle MPK is equilateral triangle

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1/5

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