Japan “Sangaku” Research Institute

Problem If we take any point $P$ on the bisector of the exterior angle of $∠A$ of $△ABC$, we get $$PB+PC>AB+AC.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution Take the point $D$ on the extension of $BA$ so…

Problem Take any point $D$ on the side $AB$ of $△ABC$, and any point $F$ on the extension of $AC$. Connecting $D$ and $F$, let $N$ be the intersection of the bisectors of $∠ADF$ and $∠ABC$, and let $M$ be the intersection of the bisectors of $∠AFD$ and $∠ACB$. Then, $$∠BND=∠CMF.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$…

Problem In a triangle $ABC$, let $D$ be the intersection of the bisector of $∠A$ and the perpendicular bisector of the side $BC$. When we draw the perpendiculars $DX$ and $DY$ from $D$ to the sides $AB$ and $AC$ or their extensions, $$AX=AY \qquad and \qquad BX=CY.$$   $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$…

Problem In a triangle $ABC$, let $X$ be the point where the bisector of $∠C$ intersects the side $AB$. Let $Y$ and $Z$ be the points where a straight line drawn parallel to the side $AC$ through $X$ intersects the side $BC$ and the bisector of the exterior angle of $∠C$, respectively. Then $$XY=YZ.$$ $$ $$…

Problem In a triangle $ABC$ such that $∠B=3∠C$, if $BD$ is the perpendicular drawn from $B$ to the bisector of $∠A$, then $$BD=\frac{1}{2} (AC-AB).$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution Let $E$ be the point where the…

Problem Let $D$ be the midpoint of the side $BC$ of a triangle $ABC$, and let $E$ and $F$ be the points where the bisectors of $∠ADB$ and $∠ADC$ intersect the sides $AB$ and $AC$, respectively. Then $$EF<BE+CF.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$…

Problem In a triangle $ABC$, if we take any point $P$ on the bisector of $∠A$, we have $$|AB-AC|>|PB-PC|.$$ However, let $AB≠AC$. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution $(1)$ Let $AB>AC$. If we take a point $D$…

Problem First, create a new triangle $A’B’C’$ by connecting the intersections of the bisectors of each exterior angle of the original triangle $ABC$. Next, create a triangle $A^”B^”C^”$ by connecting the intersections of the bisectors of each exterior angle of the triangle $A’B’C’$. If you create new triangles one after another in this way, the…

Problem In a triangle $ABC$, if $∠B>∠C$, then the bisector $BD$ of $∠B$ is smaller than the bisector $CE$ of $∠C$. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution Since $∠B>∠C$, $$∠ABD>∠ACE.$$ Now, if we take $∠DBF$ equal to…