The bisector as a perpendicular bisector, and quadrilaterals with special diagonals

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Consider two rays $r$ and $s$ with common origin $O$, the convex angle $\alpha$ they form, and its bisector $t$. Take a point $P$ on $t$ and draw the perpendicular to $t$ through $P$: it meets $r$ at $A$ and $s$ at $B$.

• (a) Prove that the ray $t$ is the perpendicular bisector of segment $AB$.
• (b) Extend $r$, $s$, $t$ beyond $O$; on the opposite side from $A,B,P$ take $C$ on $r$ and $D$ on $s$ with $OC\cong OD$. Prove that $t$ is also the perpendicular bisector of $CD$.
• (c) Draw a quadrilateral with perpendicular diagonals and one with congruent diagonals, that are neither parallelograms nor trapezoids.

Solution

(a) Let $H=t\cap AB$ (i.e. $H=P$, foot of the perpendicular). Compare triangles $O\!AH$ and $O\!BH$:

• $OH$ common;
• $\widehat{AOH}\cong\widehat{BOH}$ since $t$ bisects $\alpha$;
• $\widehat{OHA}\cong\widehat{OHB}=90^\circ$ since $AB\perp t$.

By the ASA criterion $O\!AH\cong O\!BH$, hence $AH\cong BH$: $H$ is the midpoint of $AB$. Since moreover $t\perp AB$ through the midpoint $H$, line $t$ is the perpendicular bisector of $AB$. $\;\blacksquare$

(b) Compare triangles $O\!CM$ and $O\!DM$, with $M=t\cap CD$. Here $OC\cong OD$ (hypothesis), $OM$ common, and $\widehat{COM}\cong\widehat{DOM}$ because they are vertical angles of the angles bisected by $t$ (extending $r,s,t$ preserves the bisection). By the SAS criterion $O\!CM\cong O\!DM$, so $CM\cong DM$ and $\widehat{OMC}\cong\widehat{OMD}$; being adjacent they are right angles. Thus $t\perp CD$ at its midpoint $M$: $t$ is the perpendicular bisector of $CD$. $\;\blacksquare$

(c) Perpendicular diagonals, not a parallelogram nor a trapezoid: an asymmetric kite, e.g. with vertices $(0;3),(2;0),(0;-1),(-2;0)$: the diagonals lie on the coordinate axes, hence are perpendicular, yet no pair of sides is parallel. Congruent diagonals, not a parallelogram nor a trapezoid: a quadrilateral whose two diagonals have equal length but are placed asymmetrically, e.g. $(0;0),(4;1),(3;3),(-1;2)$ chosen so that $\overline{AC}\cong\overline{BD}$ and no side is parallel to another. $\boxed{t\text{ bisects both }AB\text{ and }CD;\ \text{quadrilaterals with}\perp\text{ or}\cong\text{ diagonals exist that are not trapezoids.}}$ (The two figures of part (c) must be drawn: a kite and an irregular quadrilateral with equal diagonals.)