रचनाएँ (Constructions)

A. Perpendicular Bisector of a Line Segment $AB$:

1. With compass needle at $A$, open the compass to a radius strictly greater than half of $AB$ ($r > \frac{1}{2}AB$). Draw two arcs, one above and one below $AB$.
2. With the same radius, place the needle at $B$ and draw two arcs intersecting the previous arcs at points $P$ and $Q$.
3. Join $PQ$. The line $PQ$ bisects $AB$ at $90^\circ$ at midpoint $M$. Every point on $PQ$ is equidistant from $A$ and $B$ ($PA = PB$).

B. Angle Bisector of $\angle AOB$:

1. With vertex $O$ as center, draw an arc of any convenient radius intersecting ray $OA$ at $P$ and ray $OB$ at $Q$.
2. With $P$ as center and radius $> \frac{1}{2}PQ$, draw an arc inside the angle. With $Q$ as center and the same radius, draw another arc intersecting at $R$.
3. Join ray $OR$. The ray $OR$ bisects $\angle AOB$ into two equal halves ($\angle AOR = \angle BOR$).

A. Construction of $60^\circ$ and $120^\circ$:

1. Draw a ray $OA$. With $O$ as center, draw a large semicircle arc intersecting $OA$ at $P$.
2. With the same radius (do not change compass width!), place needle at $P$ and draw an arc cutting the semicircle at $Q$. The angle $\angle AOQ = \mathbf{60^\circ}$ (equilateral triangle principle!).
3. With needle at $Q$ and same radius, draw a second arc cutting at $R$. The angle $\angle AOR = \mathbf{120^\circ}$.

B. Construction of $90^\circ$ and $45^\circ$:

• $90^\circ$: Bisect the $60^\circ$ gap between $Q$ ($60^\circ$) and $R$ ($120^\circ$): $60^\circ + \frac{1}{2}(60^\circ) = 60^\circ + 30^\circ = \mathbf{90^\circ}$.
• $45^\circ$: Bisect the $90^\circ$ angle: $\frac{90^\circ}{2} = \mathbf{45^\circ}$.
• $30^\circ$: Bisect the $60^\circ$ angle: $\frac{60^\circ}{2} = \mathbf{30^\circ}$.
• $75^\circ$: Bisect the angle between $60^\circ$ and $90^\circ$: $60^\circ + \frac{30^\circ}{2} = \mathbf{75^\circ}$.

To construct a line parallel to a given line $l$ passing through an external point $P$ not on $l$:

1. Take any point $Q$ on line $l$. Join $P$ and $Q$.
2. With $Q$ as center and any radius, draw an arc intersecting line $l$ at $C$ and line $QP$ at $D$.
3. With $P$ as center and the exact same radius, draw an arc on the opposite side of transversal $QP$, cutting $QP$ at $E$.
4. Adjust compass width to match the chord distance $CD$. With $E$ as center, draw an arc intersecting the arc drawn from $P$ at point $F$.
5. Draw a straight line passing through points $P$ and $F$. Since the Alternate Interior Angles are equal ($\angle PQ C = \angle QPF$), line $PF$ is strictly parallel to line $l$ ($m \parallel l$).

1. 1. SSS (Three Sides Given: $a, b, c$):

   Draw base $BC = a$. With $B$ as center and radius $c$, draw an arc. With $C$ as center and radius $b$, draw an arc intersecting at $A$. Join $AB$ and $AC$. (Condition: $a+b > c$).

2. 2. SAS (Two Sides & Included Angle Given: $b, c, \angle A$):

   Draw segment $AB = c$. At vertex $A$, construct the given angle $\angle BAX = \theta$ using compasses. From ray $AX$, cut off segment $AC = b$. Join $BC$.

3. 3. ASA (Two Angles & Included Side Given: $\angle B, \angle C, a$):

   Draw base $BC = a$. At vertex $B$, construct ray making angle $\angle B$. At vertex $C$, construct ray making angle $\angle C$. The intersection of the two rays defines vertex $A$.

4. 4. RHS (Right-Angled Triangle: Hypotenuse $h$, Leg $b$):

   Draw base $BC = b$. At vertex $B$, construct a $90^\circ$ perpendicular ray $BX$. With $C$ as center and radius equal to the hypotenuse $h$, draw an arc cutting ray $BX$ at $A$. Join $AC$.