Let angle xAy be acute, B a point on ray Ax, and Bm a ray inside angle xAy with Bm parallel to Ay. Let Au and Bv bisect angle xAy and angle xBm. (a) Prove Au is parallel to Bv. (b) If Bz bisects angle ABm, prove angle vBz = 90 degrees. (c) Deduce that Bz is perpendicular to Au.

Step-by-step solution

1. 1

   (a) Transfer the angle using the parallel ray. Since $Bm\parallel Ay$ and $Ax$ is a transversal cutting both at $A$ and $B$, the angles $\angle xBm$ and $\angle xAy$ are corresponding angles, hence equal:

   $\angle xBm=\angle xAy$

2. 2

   Halve both equal angles. Bisectors cut equal angles into equal halves, so

   $\angle xBv=\tfrac12\angle xBm=\tfrac12\angle xAy=\angle xAu$

   These two are again corresponding angles for the transversal $Ax$ cutting $Au$ and $Bv$. Equal corresponding angles force the lines to be parallel:

   $Au\parallel Bv$

3. 3

   (b) Identify the supplementary pair at B. The points $A$, $B$ and the direction $x$ lie on one straight line with $B$ between $A$ and the direction of $x$, so rays $BA$ and $Bx$ are opposite. Hence $\angle ABm$ and $\angle xBm$ are a linear pair:

   $\angle ABm+\angle xBm=180^{\circ}$

4. 4

   Halve the linear pair. $Bz$ bisects $\angle ABm$ and $Bv$ bisects $\angle xBm$, and the two bisected angles are adjacent along the line, so

   $\angle vBz=\tfrac12\angle xBm+\tfrac12\angle ABm=\tfrac12\left(\angle xBm+\angle ABm\right)=\tfrac12(180^{\circ})=90^{\circ}$

   $\boxed{\angle vBz=90^{\circ}}$

   This is the general fact that the bisectors of two supplementary adjacent angles are always perpendicular.

5. 5

   (c) Combine the two results. From (b), $Bz\perp Bv$. From (a), $Bv\parallel Au$. A line perpendicular to one of two parallel lines is perpendicular to the other, so

   $Bz\perp Au$

   $\boxed{Au\parallel Bv,\quad \angle vBz=90^{\circ},\quad Bz\perp Au}$

6. 6

   Check with a concrete angle. Take $\angle xAy=60^{\circ}$. Then $\angle xBm=60^{\circ}$, so $\angle xBv=30^{\circ}=\angle xAu$ ✓ (equal corresponding angles). Also $\angle ABm=120^{\circ}$, so $\angle ABz=60^{\circ}$ and $\angle vBz=30^{\circ}+60^{\circ}=90^{\circ}$ ✓. Note the conclusion in (c) is $Bz\perp Au$, not $Bz\perp Ay$ — the two are different lines unless $Au$ happens to coincide with $Ay$.