AB + AC = 3BC, midpoints and incircle related, <BIC_1 +<CIB_1 = 180^o

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AB + AC = 3BC, midpoints and incircle related, <BIC_1 +<CIB_1 = 180^o

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Source: Tuymaada Junior 2005 p7

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#1 h May 12, 2019, 6:56 PM • 1 Y

Y by Adventure10

The point $I$ is the center of the inscribed circle of the triangle $ABC$ . The points $B_1$ and $C_1$ are the midpoints of the sides $AC$ and $AB$ , respectively. It is known that $\angle BIC_1 + \angle CIB_1 = 180^\circ$ . Prove the equality $AB + AC = 3BC$

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#2 h May 12, 2019, 8:40 PM • 3 Y

Y by Vlados021, Adventure10, Mango247

First note that if $\angle BIC_1 + \angle CIB_1 = 180^\circ$ than $\angle C_1IB_1 + \angle CIB = 180^\circ$ , so $\angle C_1IB_1 = 180^\circ -\angle CIB = \angle ABC/2 +\angle ACB/2$ so circumcircle of the triangle $BC_1I$ tangent to the circumcircle of triangle $B_1IC$ (Since $\angle C_1BI + \angle B_1CI =\angle ABC/2 +\angle ACB/2 = \angle C_1IB_1$ ). Let they (circumcircles) meet $BC$ at points $K$ and $M$ respectively.
Note that $\angle KIM =\angle ABC/2 +\angle ACB/2$ from the tangency and so $\angle KIM = \angle C_1IB_1$ . Also we see that $\angle IKC_1 = \angle IC_1K = IBK$ so $C_1I = IK$ . Similarly $IM = IB_1$ so triangle $KIM$ equal to triangle $C_1IB_1$ so $C_1B_1 = KM$ .
By the second sparrow lemma $CK = AC - AB/2$ and $BM = AB - AC/2$ so we have
$BC = BM + CK - KM = AB - AC/2 + AC - AB/2 - BC/2$ or
$AC+ AB = 3BC$ q.e.d.

This post has been edited 5 times. Last edited by Daniil02, May 12, 2019, 8:48 PM

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#3 h May 13, 2019, 5:00 PM • 3 Y

Y by Pluto1708, Adventure10, Mango247

Note that
$\angle C_1IB_1=360^\circ-\angle CIB_1-\angle BIC_1-\angle BIC=180^\circ-\angle BIC=90^\circ-\frac{\angle BAC}{2},$ which since $I$ lies on the bisector of $\angle BAC$ implies that $I$ is the $A$ -excenter of $\triangle AB_1C_1$ . It follows that $BC_1B_1C$ is circumscribed, thus
$BC_1+CB_1=B_1C_1+BC\Leftrightarrow \frac{AB}{2}+\frac{AC}{2}=\frac{BC}{2}+BC\Leftrightarrow AB+AC=3BC.$

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Plops

#4 h May 13, 2019, 11:32 PM • 1 Y

Y by Adventure10

This is very similar to CMIMC $2016$ Geometry Problem $7.$

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