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equilateral triangle criterion with distances, BB_1 = d(B_1, AB) + d(A_1, BC)

Source: 2008 Moldova JBMO TST 2.3

July 12, 2020
geometryEquilateral Triangledistance

Problem Statement

In an acute triangle ABCABC, points A1,B1,C1A_1, B_1, C_1 are the midpoints of the sides BC,AC,ABBC, AC, AB, respectively. It is known that AA1=d(A1,AB)+d(A1,AC)AA_1 = d(A_1, AB) + d(A_1, AC), BB1=d(B1,AB)+d(A1,BC)BB1 = d(B_1, AB) + d(A_1, BC), CC1=d(C1,AC)+d(C1,BC)CC_1 = d(C_1, AC) + d(C_1, BC), where d(X,YZ)d(X, Y Z) denotes the distance from point XX to the line YZYZ. Prove, that triangle ABCABC is equilateral.
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