Problem in Triangles.

Question

$A{A_1}, B{B_1}, C{C_1}$ are the median of triangle ${ABC}$ whose centroid is ${G}$ . If the points ${A}, {C_1} ,{G}, {B_1}$ are concyclic, then show that ${2a^2 =c^2 +b^2} $ if $a,b,c$ are sides of triangle?

Answer 1

Let $a$, $b$ and $c$ be sides-lengths of $\Delta ABC$ in the standard notation.

Thus, by the Ptolemy theorem we obtain $$AC_1\cdot B_1G+AB_1\cdot C_1G= C_1B_1\cdot AG$$ or $$\frac{c}{2}\cdot\frac{1}{6}\sqrt{2a^2+2c^2-b^2}+\frac{b}{2}\cdot\frac{1}{6}\sqrt{2a^2+2b^2-c^2}=\frac{a}{2}\cdot\frac{1}{3}\sqrt{2b^2+2c^2-a^2}$$ or after squaring of the both sides $$2a^4+b^4+c^4-3a^2b^2-3a^2c^2-b^2c^2+bc\sqrt{(2a^2+c^2-b^2)(2a^2+2b^2-c^2)}=0,$$ which gives $$(2a^4+b^4+c^4-3a^2b^2-3a^2c^2-b^2c^2+bc)^2-b^2c^2(2a^2+c^2-b^2)(2a^2+2b^2-c^2)=0$$ or $$(a+b+c)(a+b-c)(a+c-b)(b+c-a)(b^2+c^2-2a^2)^2=0,$$ which gives $$b^2+c^2=2a^2.$$

Answer 2

I have one more solution. Say $AA_1 = 3x$. Since $$\angle B_1BC =\angle C_1B_1B = C_1AG$$ line $BC$ is tangent on circumcircle of triangle $ABG$ so by PoP we have $$a^2/4=A_1B^2 =A_1G\cdot A_1A = 3x^2$$ Now since we have paralelogram identity we have $$(6x)^2 +a^2 = 2b^2+2c^2$$ and we are done.