Comprehensive theory, key formulas, diagrams, and memory aids for Constructions.
In geometry, "construction" means drawing shapes, angles, or lines accurately. These drawings are done using only two instruments: an ungraduated ruler (straightedge) and a compass. While a protractor and set squares are useful for quick drawings, true Euclidean constructions rely solely on the pure geometric properties of circles (drawn with a compass) and straight lines.
In this chapter, we will look at basic constructions and how to construct specific triangles given certain conditions.
Before constructing complex triangles, you must master the fundamental constructions.
Objective: To bisect (cut exactly in half) a given angle $\angle ABC$. Steps of Construction: 1. Taking $B$ as the centre and any convenient radius, draw an arc intersecting the rays $BA$ and $BC$, say at $E$ and $D$ respectively. 2. Taking $D$ and $E$ as centres and with the radius more than $\frac{1}{2} DE$, draw arcs to intersect each other, say at $F$. 3. Draw the ray $BF$. This ray $BF$ is the required bisector of the angle $ABC$.
Why it works: By joining $DF$ and $EF$, we form two triangles, $\triangle BEF$ and $\triangle BDF$. $BE=BD$ (radii of same arc), $EF=DF$ (radii of same arcs), and $BF$ is common. By SSS congruence, the triangles are congruent, meaning $\angle EBF = \angle DBF$.
Objective: To draw a line that cuts a given line segment $AB$ exactly in half at a $90^\circ$ angle. Steps of Construction: 1. Taking $A$ and $B$ as centres and radius more than $\frac{1}{2} AB$, draw arcs on both sides of the line segment $AB$. 2. Let these arcs intersect each other at $P$ and $Q$. Join $PQ$. 3. Let $PQ$ intersect $AB$ at the point $M$. Then line $PMQ$ is the required perpendicular bisector of $AB$.
Why it works: Joining $A$ and $B$ to $P$ and $Q$ forms $\triangle PAQ$ and $\triangle PBQ$. By SSS congruence ($AP=BP$, $AQ=BQ$, $PQ$ common), they are congruent, making $\angle APM = \angle BPM$. Then, in $\triangle PMA$ and $\triangle PMB$, $AP=BP$, $PM$ is common, and the included angle is equal. By SAS, they are congruent. This makes $AM=MB$ (bisector) and $\angle PMA = \angle PMB = 90^\circ$ (perpendicular).
Steps of Construction: 1. Draw a ray $AB$ with initial point $A$. 2. Taking $A$ as the centre and some radius, draw an arc of a circle, which intersects $AB$, say at a point $D$. 3. Taking $D$ as the centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point $E$. 4. Draw the ray $AC$ passing through $E$. Then $\angle CAB = 60^\circ$.
Why it works: If you join $DE$, you form a triangle $ADE$. Since $AD = AE = DE$ (all drawn with the same radius), $\triangle ADE$ is an equilateral triangle. In an equilateral triangle, all interior angles are $60^\circ$.
By repeatedly bisecting angles, you can use the $60^\circ$ construction to make $30^\circ, 15^\circ, 90^\circ, 45^\circ$, and so on.
To construct a unique triangle, we need three independent measurements (like three sides, or two sides and an included angle). However, sometimes we are given combinations of sides and angles.
Objective: Construct a triangle $ABC$ given base $BC$, a base angle $\angle B$, and the sum of other two sides $AB + AC$. Steps of Construction: 1. Draw the base $BC$ and at the point $B$ make an angle, say $XBC$ equal to the given angle. 2. Cut a line segment $BD$ equal to $AB + AC$ from the ray $BX$. 3. Join $DC$ and make an angle $\angle DCY$ equal to $\angle BDC$. 4. Let $CY$ intersect $BX$ at $A$. Then, $\triangle ABC$ is the required triangle.
Alternative Method for step 3 & 4: Instead of drawing the angle, you can draw the perpendicular bisector of $CD$. Let it intersect $BD$ at point $A$. Join $AC$. Because $A$ lies on the perpendicular bisector of $CD$, $AD = AC$. Thus $AB = BD - AD = BD - AC$, which means $AB + AC = BD$.
Objective: Construct $\triangle ABC$ given base $BC$, base angle $\angle B$, and the difference of other two sides ($AB - AC$ or $AC - AB$).
Case 1: $AB > AC$ (i.e., $AB - AC$ is given) 1. Draw the base $BC$ and at point $B$ make an angle say $XBC$ equal to the given angle. 2. Cut the line segment $BD$ equal to $AB - AC$ from ray $BX$. 3. Join $DC$ and draw the perpendicular bisector of $DC$. 4. Let it intersect $BX$ at a point $A$. Join $AC$. $\triangle ABC$ is the required triangle. (Proof: $A$ is on perpendicular bisector of $CD$, so $AD = AC$. $AB = BD + AD = BD + AC$, so $BD = AB - AC$).
Case 2: $AB < AC$ (i.e., $AC - AB$ is given) 1. Draw the base $BC$ and at point $B$ make an angle $XBC$. 2. Extend the ray $XB$ backwards to $X'$. 3. Cut the line segment $BD$ equal to $AC - AB$ from the extended line $BX'$ (below base $BC$). 4. Join $DC$ and draw the perpendicular bisector of $DC$. 5. Let it intersect $BX$ at $A$. Join $AC$. $\triangle ABC$ is the required triangle.
Objective: Construct a triangle given its perimeter ($AB + BC + CA$) and its two base angles, say $\angle B$ and $\angle C$. Steps of Construction: 1. Draw a line segment, say $XY$, equal to $AB + BC + CA$. 2. Make angles $\angle LXY$ equal to $\angle B$ and $\angle MYX$ equal to $\angle C$. 3. Bisect $\angle LXY$ and $\angle MYX$. Let these bisectors intersect at a point $A$. 4. Draw perpendicular bisectors $PQ$ of $AX$ and $RS$ of $AY$. 5. Let $PQ$ intersect $XY$ at $B$ and $RS$ intersect $XY$ at $C$. Join $AB$ and $AC$. 6. Then $\triangle ABC$ is the required triangle.
Geometric constructions rely on the pure logic of congruency. By understanding why these construction steps work (usually rooted in SSS, SAS, and properties of perpendicular bisectors), you can figure out how to construct almost any shape given sufficient constraints.