Let D be the midpoint of the side BC of a tri | vedclass

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(B) Let the side length of the equilateral triangle $ADC$ be $x$. Thus,$DC = AD = AC = x$.Since $D$ is the midpoint of $BC$,$BD = DC = x$,so $BC = a =

Vedclass · Accepted Answer

$4:1:3$

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(B) Let the side length of the equilateral triangle $ADC$ be $x$. Thus,$DC = AD = AC = x$.Since $D$ is the midpoint of $BC$,$BD = DC = x$,so $BC = a = 2x$.In $	riangle ADC$,all angles are $60^\circ$. Since $\angle ADC = 60^\circ$,the adjacent angle $\angle ADB = 180^\circ - 60^\circ = 120^\circ$.In $	riangle ABD$,by the Law of Cosines:$AB^2 = BD^2 + AD^2 - 2(BD)(AD) \cos(120^\circ)$$c^2 = x^2 + x^2 - 2(x)(x)(-1/2)$$c^2 = 2x^2 + x^2 = 3x^2$.We have $a = 2x$,$b = AC = x$,and $c^2 = 3x^2$.Therefore,$a^2 : b^2 : c^2 = (2x)^2 : x^2 : 3x^2 = 4x^2 : x^2 : 3x^2 = 4:1:3$.

Let $D$ be the midpoint of the side $BC$ of a triangle $ABC$. If the triangle $ADC$ is equilateral,then the ratio $a^2 : b^2 : c^2$ is equal to

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