Japan “Sangaku” Research Institute

Problem Let $D$ and $E$ be the feet of the perpendiculars drawn from $C$ to the bisectors of $∠A$ of $△ABC$ and its exterior angle, respectively, and let $M$ be the midpoint of the side $BC$. Then, $M, \ D$ and $E$ are on the same straight line. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$…

Problem Let $P$ and $Q$ be the feet of the perpendiculars drawn from $A$ to the bisectors of $∠B$ and $∠C$ of $△ABC$, respectively. Then, $$PQ=\frac{1}{2} (AB+AC-BC).$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution Let the intersections of…

Problem Let $O$ be the intersection of the bisectors of $∠B$ and $∠C$ of $△ABC$, and let $D$ and $E$ be the feet of the perpendicular lines drawn from $A$ to the straight lines $BO$ and $CO$. Then $$DE∥BC.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$…

Problem In $△ABC$, if we draw perpendiculars $BG$ and $CH$ from $B$ and $C$ to the bisector of the exterior angle of $∠A$, and let $AD$ be the bisector of $∠A$, then $CG, \ BH$ and $AD$ pass through the same point. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$…

Problem Let $E$ and $F$ be the feet of the perpendiculars drawn from both ends $B$ and $C$ of the base to the bisector of the apex angle $∠A$ of $△ABC$, and let $G$ be the midpoint of $BC$. Then, $△GEF$ is an isosceles triangle. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$…

Problem Suppose that the side $AB$ of $△ABC$ is one third of the side $AC$. Drop the perpendicular line $CF$ from $C$ to the bisector $AD$ of $∠A$. Then, $$AD=DF.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution Let…

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, $$AX=AY \qquad and \qquad BX=CY.$$   $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$…