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In a triangle ABC, the angle BAC is equal to 120 degrees. If AB = c, AC = b, and AD = x, and x is the angle bisector of angle BAC, prove that 1/x=1/b+1/c

 Feb 11, 2023
 #3
avatar+132 
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Ok i figured it out. For anyone curious, draw an equilateral triangle in triangle ACD by drawing a parallel line to AB. Label the point E so you have an equilateral triangle ADE. Notice triangle CDE and ABC are similar. And see what you can do from there :)
 Feb 11, 2023
 #4
avatar+132 
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If we draw a parallel line to $AB$ and mark the intersection point $E$, we create an equilateral $\triangle ADE$. This means we can derive the following information: $$a_1 = AD, DE, AE$$ $$\triangle CDE ~ \triangle ABC*$$ $$\frac{DE}{AB}=\frac{CE}{AC}**$$ *by AA similarity ($120^{\circ}$ and $\angle C$) **by rules for similar triangles Notice in $\frac{DE}{AB}=\frac{CE}{AC}$ we can substitute $DE, AB, CE,$ and $AC$ with values of $a_1, b_1,$ and $c_1$. $$DE = a_1$$ $$AB = c_1$$ $$CE = b_1 - a_1$$ $$AC = b_1$$ Substituting those values, we can say: $$\frac{a}{c}=\frac{b-a}{b}$$ Multiplying both sides by $bc$ to eliminate the fraction, we get: $$ab = bc -ac$$ Add $ac$ to both sides: $$ab + ac = bc$$ Factor $a$ from the left-hand side: $$a(b+c)=bc$$ Divide both sides by $(b+c)$ and we get: $$a=\frac{bc}{b+c}$$ We can take the reciprocal of both sides to get: $$\frac{1}{a} = \frac{b+c}{bc}$$ Finally, we get: $$\frac{1}{a} = \frac{1}{b} + \frac{1}{c}$$

https://dochub.com/axeljump6/275eAYrVo3Azqk6VzXnBNQ/img-9763-jpg?dt=uysQgkj4EtEJrsHWdcpV
Saphia1123  Feb 11, 2023

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