angle bisector // MN, BD = CE. midpoints (Argentina 2019 OMA L1 p3)

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angle bisector // MN, BD = CE. midpoints (Argentina 2019 OMA L1 p3)

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#1 h May 15, 2020, 11:51 PM

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In triangle $ABC$ let $D$ and $E$ be on sides $AB$ and $AC$ , respectively, such that $BD = CE$ . Let $M$ and $N$ be the midpoints of $BC$ and $DE$ , respectively. Prove that the bisector of the angle $\angle BAC$ is parallel to the line $MN$ .

This post has been edited 1 time. Last edited by parmenides51, Jan 26, 2025, 7:05 AM

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#2 h May 18, 2020, 3:46 PM

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Let $A(0,0),B(a,b),C(c,0),E(c-\lambda,0)$ .
Line $AB\ :\ y=\frac{b}{a}x$ with point $D(a-\frac{a\lambda}{\sqrt{a^{2}+b^{2}}},b-\frac{b\lambda}{\sqrt{a^{2}+b^{2}}})$ , so $BD=\lambda$ .

Midpoints $M(\frac{a+c}{2},\frac{b}{2})$ and $N(\frac{a+c-\lambda-\frac{a\lambda}{\sqrt{a^{2}+b^{2}}}}{2},\frac{b-\frac{b\lambda}{\sqrt{a^{2}+b^{2}}}}{2})$ .
Slope of the line $MN\ :\ m_{MN}=\frac{b}{a+\sqrt{a^{2}+b^{2}}}$ .

Equation of the bisector of the angle $\angle BAC$ is $y(a+\sqrt{a^{2}+b^{2}})=bx$ .
Conclusion: $MN //\$ the bisector of the angle $\angle BAC$ .

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#3 h May 18, 2020, 8:07 PM

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Draw $C'$ , $E'$ so $AC' = AB$ , $AE'=AD$ ; label midpoints of $BC'$ , $DE'$ as $M'$ , $N'$ .
Based on connecting midpoints $NN' ||EE'$ and $MM' || CC'$ .
$EE'$ , $CC'$ have length $AC-AB$ and $MM'$ , $NN'$ have half that length.
Then $M'N'$ is parallel to $MN$ as each point is equal distance away along a parallel line.
$M'N'$ is <bisector as each is midpoint of base of isosceles triangle with $<BAC$ .

This post has been edited 9 times. Last edited by EFrame2, May 18, 2020, 8:41 PM

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#4 h Jan 26, 2025, 1:53 AM

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In triangle $ABC$ let $D$ and $E$ be on sides $AB$ and $AC$ , respectively, such that $BD = CE$ . Let $M$ and $N$ be the midpoints of $BC$ and $DE$ , respectively. Prove that the bisector of the angle $\angle BAC$ is parallel to the line $MN$ .

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#5 h Jan 29, 2025, 7:04 AM

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Remark: Call $O, O'$ the circumcenters of triangles $ABC, ADE$ ; then $OO'$ is parallel to the internal bisector of $\angle BAC$ !.

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Sunken rock

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