Problem E:15741 GMB 6-7-8/2020, pag 367

 

 Proposed for the sixth grade. In translation:

        " Consider the triangle ABC with AB < AC and T a point on the line AC so that A is between C and T. The bisector of the angle TAB intersects the line BC in M, and the bisector of the angle BAC intersects the side BC in D. The point N is the symmetry of the point M with respect to the point A and E is the intersection of the lines AC and DN. Show that BE$\parallel$MN"

 

Solution CiP

 

           The figure in the problem looks like this:

 

          We know that the inner bisector $[AD$ and the outer bisector $[AM$ are perpendicular.

 

(Because $\widehat{BAD}+\widehat{BAM}=\frac{\widehat{BAC}}{2}+\frac{\widehat{BAT}}{2}=\frac{\widehat{BAC}}{2}+\frac{180^{\circ}-\widehat{BAC}}{2}=90^{\circ}$)

           In the figure of the problem, point $A$ is the middle of the segment $[MN]$ and $DA\perp MA$, so we have the relation between lines:

$(1)$                                        $AD\perp MN$.

It follows that the line $AD$ is the mediator of the segment $[MN]$. In the isosceles triangle $DMN$, of vertex $D$ and the base$[ MN]$, the mediator $AD$ is also bisectors of the angle from the vertex, $\sphericalangle MDN$, so

$(2)$                         $\sphericalangle ADM \equiv \sphericalangle ADN$.

           From $\sphericalangle BAD \equiv \sphericalangle CAD$ (see hypothesis), $\sphericalangle ADB \equiv  \sphericalangle ADE$ (see (2)) we obtain the congruence of the triangles with the common side $[AD]$

$\triangle ABD \equiv \triangle AED$ (ASA rule).

 From here results $[AB] \equiv [AE]$, and in the isosceles triangle $ABE$ the bisector at the top $AD$ is also height, so

$(3)$                         $AD \perp BE$.

      The lines $BE$ and $MN$ are both perpendicular to the line $AD$ (see (1), (3)) and hence $BE \parallel MN$.

$\blacksquare$