Question

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.

Answer 1

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