Construction of Triangles: Advanced Cases

Construction of a Triangle when Two Sides and Median is Given

This construction problem requires building a triangle when you know the lengths of two of its sides and the length of the median drawn from the vertex connecting those two sides to the opposite side.

Let the triangle be $\triangle ABC$. Suppose the given sides are $AB=c$ and $AC=b$, and the given median is $AD=m_a$, where $D$ is the midpoint of $BC$. The strategy is to construct an auxiliary figure, usually a parallelogram, that incorporates the given lengths in a way that allows the triangle to be formed.

Construction Steps

Given: Lengths of two sides, $b$ and $c$, and the length of the median to the third side (from the vertex where sides $b$ and $c$ meet), $m_a$. Let's construct $\triangle ABC$ with $AC=b$, $AB=c$, and median $AD=m_a$ to side $BC$.

Tools: Ruler, Compass, Straightedge.

Goal: Construct $\triangle ABC$.

Auxiliary Construction Idea: Imagine duplicating the median $AD$ to a point $E$ such that $AD = DE$ and $A, D, E$ are collinear. If we then form quadrilateral $ABEC$, the diagonals $AE$ (length $2m_a$) and $BC$ will bisect each other at $D$. A quadrilateral whose diagonals bisect each other is a parallelogram. In parallelogram $ABEC$, opposite sides are equal, so $AB=EC=c$ and $AC=BE=b$. This suggests we can construct a triangle using sides $b, c,$ and $2m_a$.

Steps:

1. Construct Auxiliary Triangle: Construct a triangle, say $\triangle ACE'$, using the SSS criterion (I1). The sides of this auxiliary triangle will have the following lengths:
   • Side $AC'$ = $b$ (the length of one given side)
   • Side $CE'$ = $c$ (the length of the other given side)
   • Side $AE'$ = $2m_a$ (twice the length of the given median)
   (Important Check: Before constructing, ensure that the Triangle Inequality Theorem holds for the sides $b, c, 2m_a$. If $b+c \le 2m_a$, $b+2m_a \le c$, or $c+2m_a \le b$, this auxiliary triangle cannot be formed, and thus the original triangle cannot be constructed with the given dimensions.)
2. Locate the Midpoint: The segment $AE'$ in the auxiliary triangle represents $2m_a$. Find the midpoint of the side $AE'$. This can be done by constructing the perpendicular bisector of $AE'$ (I3) to find its midpoint, let's call it $D$. Since $AE' = 2m_a$, $AD = DE' = m_a$.
3. Locate Vertex B: The point $D$ is the midpoint of the median $AE'$ (which corresponds to $AD$ in the original triangle). In our conceptual parallelogram $ABEC$, $D$ is also the midpoint of $BC$. We need to find vertex $B$. Join point $C$ to point $D$. Extend the line segment $CD$ beyond $D$. Using a compass, measure the length of $CD$. Place the compass point on $D$ and mark off a point $B$ on the extended ray from $C$ through $D$ such that $DB = CD$.
4. Complete the Triangle: The vertices of the required triangle are $A$, $B$, and $C$. Join point $A$ to point $B$ using a straightedge. Point $A$ and $C$ are already joined as a side of the auxiliary triangle.

The resulting figure $\triangle ABC$ is the required triangle.

Justification:

By construction, $AC = b$. In the auxiliary triangle $\triangle ACE'$, $CE' = c$ and $AE' = 2m_a$. $D$ is the midpoint of $AE'$, so $AD = m_a$. Vertex $B$ is located on the ray $CD$ extended such that $CD=DB$. This means $D$ is the midpoint of the segment $BC$. Therefore, $AD$ is a median to side $BC$ in $\triangle ABC$, and its length is $m_a$, as required.

Now, consider the quadrilateral $ABEC'$ (relabeling $E'$ as $E$ for clarity, where $AE=2m_a$). The diagonals $AE$ and $BC$ intersect at $D$, and by construction, $D$ is the midpoint of both $AE$ ($AD=DE=m_a$) and $BC$ ($CD=DB$). A quadrilateral whose diagonals bisect each other is a parallelogram. Therefore, quadrilateral $ABEC$ is a parallelogram.

In a parallelogram, opposite sides are equal in length. Thus, $AB = EC$ and $AC = BE$.

We constructed the auxiliary triangle $\triangle ACE$ with sides $AC=b$, $CE=c$. From the parallelogram property, $AB = CE$. Since $CE = c$, we have $AB = c$. Also, $AC=b$ is given. We constructed $AD$ as the median of length $m_a$.

So, $\triangle ABC$ has sides $AC=b$, $AB=c$, and median $AD=m_a$. This matches the given information, confirming the construction is correct.

Example

Answer:

Construction of a Triangle when Two angles and Altitude is Given

This construction creates a triangle when you are given the measures of two angles and the length of an altitude. There are multiple scenarios depending on which angles and which altitude are given. A common case is being given two base angles and the altitude from the third vertex to the base.

Let the triangle be $\triangle ABC$. Suppose the given angles are $\angle B$ and $\angle C$, and the given altitude is $h_a$ from $A$ to $BC$. The strategy often involves using the altitude to define the height of the triangle and then using angle properties in right-angled sub-triangles formed by the altitude.

Construction Steps (Altitude from the third vertex $A$ to side $BC$)

Given: Measures of two base angles, say $\angle B$ and $\angle C$, and the length of the altitude from the third vertex $A$ to the base $BC$, say $h_a$.

Condition for Construction: For a triangle to exist, $\angle B + \angle C < 180^\circ$. Also, $\angle B$ and $\angle C$ must be acute if $D$ lies between $B$ and $C$. If one angle is obtuse (say $\angle C$), the foot of the altitude $D$ will fall outside the segment $BC$. The construction needs slight modification in step 4/5 depending on which angles are given and which one is obtuse. Assuming $\angle B$ and $\angle C$ are acute for the standard construction shown.

Tools: Ruler, Compass, Straightedge, Protractor (or methods for specific constructible angles).

Goal: Construct $\triangle ABC$ with base angles $\angle B, \angle C$ and altitude $h_a$ from $A$ to $BC$.

Steps:

1. Draw a Base Line: Draw a long horizontal line $l$. This line will contain the base $BC$.
2. Construct the Altitude: Choose any point $D$ on line $l$. Construct a line perpendicular to $l$ at point $D$ (I1). On this perpendicular line, measure and mark off a segment $AD$ equal to the given altitude $h_a$. Point $A$ is the location of the third vertex.
3. Draw a Parallel Line (Optional but helpful visualization): Draw a line $m$ passing through point $A$ and parallel to line $l$ (I1). This line $m$ represents the locus of all possible positions for vertex $A$ given the altitude $h_a$ to line $l$.
4. Locate Vertex B: Now we need to find point $B$ on line $l$. We know $\angle B$ and the right angle at $D$. In the right-angled triangle $\triangle ADB$, $\angle DAB = 90^\circ - \angle B$. Construct an angle equal to $90^\circ - \angle B$ at point $A$ such that one arm is $AD$ and the other arm intersects line $l$. The arm should be on the side of $AD$ where $B$ is expected. Label the intersection point on line $l$ as $B$.
5. Locate Vertex C: Similarly, in the right-angled triangle $\triangle ADC$, $\angle DAC = 90^\circ - \angle C$. Construct an angle equal to $90^\circ - \angle C$ at point $A$ such that one arm is $AD$ and the other arm intersects line $l$. This arm should be on the opposite side of $AD$ from where $B$ was located (assuming $B$ and $C$ are on opposite sides of $D$). Label the intersection point on line $l$ as $C$.
6. Complete Triangle: Join points $A$ to $B$ and $A$ to $C$ using a straightedge (if $AB$ and $AC$ are not already drawn as rays).

The resulting figure $\triangle ABC$ is the required triangle.

Justification:

By construction, $AD$ is perpendicular to $BC$ (which lies on line $l$) and $AD = h_a$, so $AD$ is the altitude from $A$ to $BC$ with the correct length. Points $B$ and $C$ are located on line $l$.

Consider the right-angled triangle $\triangle ADB$. We constructed $\angle DAB = 90^\circ - \angle B$. The sum of angles in $\triangle ADB$ is $180^\circ$.

Thus, $\angle ABC = \angle B$, as required.

Similarly, consider the right-angled triangle $\triangle ADC$. We constructed $\angle DAC = 90^\circ - \angle C$.

Thus, $\angle ACB = \angle C$, as required.

The triangle $ABC$ has base angles $\angle B$ and $\angle C$ and altitude $h_a$ from $A$ to $BC$.

Alternative Construction (Using Base Angles directly):

1. Draw a line $l$. Construct a line $m$ parallel to $l$ at a distance $h_a$. (Draw a perpendicular from a point on $l$, mark $h_a$, draw parallel through that point). Any point on $m$ can be vertex $A$.
2. Choose any point $A$ on line $m$.
3. From $A$, draw a ray $AX$ such that the angle between $AX$ and the perpendicular from $A$ to $l$ is $90-\angle B$. Ray $AX$ intersects $l$ at $B$. (Or draw a ray from $A$ 'pointing towards' the base and construct $\angle B$ below line $m$ such that one arm is $AB$ and the other is on line $l$).
4. From $A$, draw a ray $AY$ such that the angle between $AY$ and the perpendicular from $A$ to $l$ is $90-\angle C$ (on the other side). Ray $AY$ intersects $l$ at $C$. (Or construct $\angle C$ below line $m$ from $A$).
5. Join $A$ to $B$ and $A$ to $C$.

This alternative is conceptually similar but might be harder to execute if $\angle B$ or $\angle C$ are not acute, as you need to be careful with angle directions. The first method (using angles $90-\angle B$ and $90-\angle C$ at A) is more standard.

Example

Answer:

Construction of a Triangle where One Side, One Angle and Sum of Other Two Sides are Provided

This is a specific construction problem where you are given the length of one side (usually taken as the base), the measure of one angle adjacent to that base, and the sum of the lengths of the other two sides.

Let the triangle be $\triangle ABC$. Suppose you are given base $BC=a$, angle $\angle B$, and the sum $AB+AC=s$. The strategy is to use the sum of the sides to locate a point and then use the property of perpendicular bisectors to find the third vertex.

Construction Steps

Given: Length of the base side $BC = a$, measure of the base angle $\angle B$ (adjacent to $BC$), and the sum of the lengths of the other two sides $AB + AC = s$.

Condition for Construction: For a triangle to exist, the sum of any two sides must be greater than the third. In particular, $s > a$. Also, $s$ should be greater than the segment formed by the angle and base, which is implicitly handled by the construction. The given angle $\angle B$ must be less than $180^\circ$.

Tools: Ruler, Compass, Straightedge, Protractor (or methods for specific angles).

Goal: Construct $\triangle ABC$ with $BC=a$, $\angle B$, and $AB+AC=s$.

Steps:

1. Draw the Base: Use the ruler to draw the base line segment $BC$ with length equal to $a$.
2. Construct the Base Angle: At vertex $B$ (one endpoint of the base), construct the given angle $\angle B$. Draw a ray $BX$ starting from $B$ such that $\angle CBX$ is equal to the given measure of $\angle B$. This ray will contain the side $AB$.
3. Mark the Sum Length: On the ray $BX$ (starting from $B$), measure and cut off a line segment $BD$ whose length is equal to the given sum $s$ (i.e., set compass to $s$, place at $B$, draw arc intersecting $BX$). Mark the point of intersection as $D$. Note that $BD = s = AB + AC$.
4. Join D to C: Use the straightedge to draw the line segment connecting point $D$ to point $C$.
5. Construct the Perpendicular Bisector: Construct the perpendicular bisector of the line segment $DC$ (I3). Every point on this bisector is equidistant from $D$ and $C$.
6. Locate Vertex A: The third vertex $A$ of the triangle must satisfy $AB + AC = s$. It must also lie on the ray $BX$. We created $BD$ on $BX$ such that $BD=s$. If $A$ is on $BX$, then $BD = BA + AD$ (since $A$ will lie between $B$ and $D$ if $AC < AD + CD$). We need $AB+AC = AB+AD = s$, which means $AC = AD$. The locus of points equidistant from $D$ and $C$ is the perpendicular bisector of $DC$. Therefore, vertex $A$ must be the point where the perpendicular bisector of $DC$ intersects the ray $BX$. Label this intersection point as $A$.
7. Complete the Triangle: Use the straightedge to draw the line segment connecting point $A$ to point $C$.

The resulting figure $\triangle ABC$ is the required triangle.

Justification:

By construction, the base $BC$ has length $a$ and the angle $\angle B$ has the given measure. Point $A$ lies on the ray $BX$, which was constructed at angle $\angle B$ to $BC$. The point $A$ was found as the intersection of ray $BX$ and the perpendicular bisector of $DC$. By the property of the perpendicular bisector, any point on it is equidistant from the endpoints $D$ and $C$. Therefore, $AD = AC$.

Now consider the length of the segment $BD$. By construction, $A$ lies on $BD$ (specifically, between $B$ and $D$, which will always happen if $s>a$ and $\angle B$ is acute or right). So, $BD = AB + AD$. Substituting $AD = AC$ into this equation, we get $BD = AB + AC$.

By construction (Step 3), the length $BD$ was made equal to the given sum $s$. Therefore, $AB + AC = s$.

The triangle $ABC$ constructed has side $BC = a$, angle $\angle B$, and the sum of the other two sides $AB + AC = s$. This matches all the given conditions.

Example

Answer:

Construction of a Triangle where One Side, One Angle and Difference of Other Two Sides are Provided

This construction is similar to the previous one but involves the difference between the lengths of the other two sides instead of their sum. Again, you are typically given the base, one base angle, and the difference.

Let the triangle be $\triangle ABC$. Suppose you are given base $BC=a$, angle $\angle B$, and the difference between sides $AB$ and $AC$. There are two cases depending on which of $AB$ or $AC$ is larger.

The strategy involves marking off the difference on the arm of the given angle and then using a perpendicular bisector.

Case 1: Side Adjacent to the Angle is Longer ($AB > AC$)

Given: Length of the base side $BC = a$, measure of the base angle $\angle B$ (adjacent to $BC$), and the difference $AB - AC = d$ (where $d > 0$).

Condition for Construction: For a triangle to exist, $AB - AC < BC$ by the triangle inequality ($AB < AC + BC \Rightarrow AB - AC < BC$). So, $d < a$. Also, the given angle $\angle B$ must be less than $180^\circ$.

Tools: Ruler, Compass, Straightedge, Protractor (or methods for specific angles).

Goal: Construct $\triangle ABC$ with $BC=a$, $\angle B$, and $AB - AC = d$.

Steps:

1. Draw the Base: Use the ruler to draw the base line segment $BC$ with length equal to $a$.
2. Construct the Base Angle: At vertex $B$, construct the given angle $\angle B$. Draw a ray $BX$ starting from $B$ such that $\angle CBX = \angle B$. This ray will contain the side $AB$.
3. Mark the Difference Length: On the ray $BX$ (starting from $B$), measure and cut off a line segment $BD$ whose length is equal to the given difference $d$ (i.e., set compass to $d$, place at $B$, draw arc intersecting $BX$). Mark the point of intersection as $D$. Note that $BD = d = AB - AC$. Since $AB > AC$, $D$ will be between $A$ and $B$.
4. Join D to C: Use the straightedge to draw the line segment connecting point $D$ to point $C$.
5. Construct the Perpendicular Bisector: Construct the perpendicular bisector of the line segment $DC$ (I3).
6. Locate Vertex A: The third vertex $A$ of the triangle must satisfy $AB - AC = d$. We have $BD=d$, so we need $AB - AC = BD$. From the figure, $AB = AD + DB$, so $AD + DB - AC = DB \Rightarrow AD = AC$. The locus of points equidistant from $D$ and $C$ is the perpendicular bisector of $DC$. Therefore, vertex $A$ must be the point where the perpendicular bisector of $DC$ intersects the ray $BX$. Label this intersection point as $A$. Since $d
7. Complete Triangle: Use the straightedge to draw the line segment connecting point $A$ to point $C$.

The resulting figure $\triangle ABC$ is the required triangle.

Justification:

By construction, $BC=a$ and $\angle B$ is the required angle. Point $A$ lies on the ray $BX$ and on the perpendicular bisector of $DC$. Because $A$ is on the perpendicular bisector of $DC$, $AD = AC$.

Since $A$ is on ray $BX$ and $D$ was marked on $BX$ with $BD=d$, and $AB > AC$ ($d>0$), $D$ must lie between $B$ and $A$. Thus, $AB = BD + DA$. Substituting $AD=AC$ and $BD=d$, we get $AB = d + AC$, which rearranges to $AB - AC = d$.

The triangle $ABC$ constructed has side $BC = a$, angle $\angle B$, and $AB - AC = d$. This matches the given conditions.

Case 2: Side Opposite to the Angle is Longer ($AC > AB$)

Given: Length of the base side $BC = a$, measure of the base angle $\angle B$ (adjacent to $BC$), and the difference $AC - AB = d$ (where $d > 0$).

Condition for Construction: For a triangle to exist, $AC - AB < BC$ by the triangle inequality ($AC < AB + BC \Rightarrow AC - AB < BC$). So, $d < a$. Also, the given angle $\angle B$ must be less than $180^\circ$.

Tools: Ruler, Compass, Straightedge, Protractor (or methods for specific angles).

Goal: Construct $\triangle ABC$ with $BC=a$, $\angle B$, and $AC - AB = d$.

Steps:

1. Draw the Base: Draw the base line segment $BC$ with length equal to $a$.
2. Construct the Base Angle: At vertex $B$, construct the given angle $\angle B$. Draw a ray $BX$ such that $\angle CBX = \angle B$. This ray will contain the side $AB$.
3. Extend Ray Backwards: Extend the ray $XB$ backwards beyond $B$ to create a line.
4. Mark the Difference Length: On the extended part of the line (on the side opposite to ray $BX$), measure and cut off a line segment $BD$ whose length is equal to the given difference $d$ (i.e., set compass to $d$, place at $B$, draw arc intersecting the extended ray). Mark the point of intersection as $D$. Note that $BD = d = AC - AB$. Also note that $D, B, X$ are collinear in that order.
5. Join D to C: Use the straightedge to draw the line segment connecting point $D$ to point $C$.
6. Construct the Perpendicular Bisector: Construct the perpendicular bisector of the line segment $DC$ (I3).
7. Locate Vertex A: The third vertex $A$ must satisfy $AC - AB = d$, which means $AC = AB + d$. We constructed $BD=d$. We need $AC = AB + BD$. From the figure, $AD = AB + BD$. So we need $AC = AD$. The locus of points equidistant from $D$ and $C$ is the perpendicular bisector of $DC$. Therefore, vertex $A$ must be the point where the perpendicular bisector of $DC$ intersects the original ray $BX$. Label this intersection point as $A$. Since $AC > AB$, $A$ will be on the ray $BX$, but $B$ will be between $D$ and $A$.
8. Complete Triangle: Use the straightedge to draw the line segment connecting point $A$ to point $C$.

The resulting figure $\triangle ABC$ is the required triangle.

Justification:

By construction, $BC=a$ and $\angle B$ is the required angle. Point $A$ lies on the ray $BX$ and on the perpendicular bisector of $DC$. Because $A$ is on the perpendicular bisector of $DC$, $AD = AC$.

Since $A$ is on ray $BX$ and $D$ is on the extension of $XB$, $B$ must lie between $D$ and $A$. Thus, $AD = AB + BD$. Substituting $AD=AC$ and $BD=d$, we get $AC = AB + d$, which rearranges to $AC - AB = d$.

The triangle $ABC$ constructed has side $BC = a$, angle $\angle B$, and $AC - AB = d$. This matches the given conditions.

Example

Answer:

Answer:

Construction of a Triangle where Two Angles and Sum of all the Three Sides are Provided (Perimeter)

This construction creates a triangle when you are given the measures of two angles (usually the base angles) and the perimeter (the sum of the lengths of all three sides).

Let the triangle be $\triangle ABC$. Suppose you are given angles $\angle B$ and $\angle C$, and the perimeter $P = AB + BC + AC$. The strategy involves laying out the entire perimeter as a line segment and then using properties of isosceles triangles formed by angle bisectors.

Construction Steps

Given: Measures of two angles, say $\angle B$ and $\angle C$, and the perimeter $P = AB + BC + AC$.

Condition for Construction: For a triangle to exist, $\angle B + \angle C < 180^\circ$. Also, $\angle B/2 + \angle C/2 < 180^\circ$, which is always true if $\angle B + \angle C < 360^\circ$.

Tools: Ruler, Compass, Straightedge, Protractor (or methods for specific constructible angles), Angle Bisector construction (I3).

Goal: Construct $\triangle ABC$ with angles $\angle B, \angle C$ and perimeter $P$.

Steps:

1. Draw the Perimeter Segment: Use the ruler to draw a line segment $XY$ whose length is equal to the given perimeter $P$. This segment will form the extended base containing segments equal to $AB, BC,$ and $AC$.
2. Construct Half of Base Angles:
   • At point $X$ (one endpoint of the perimeter segment), construct an angle equal to half of the given angle $\angle B$. Draw a ray $XL$ starting from $X$ such that $\angle YXL = \frac{1}{2}\angle B$. Use protractor or angle bisection (I3) if $\angle B$ is constructible.
   • At point $Y$ (the other endpoint of the perimeter segment), construct an angle equal to half of the given angle $\angle C$. Draw a ray $YM$ starting from $Y$ such that $\angle XYM = \frac{1}{2}\angle C$. Ensure this ray $YM$ is on the same side of $XY$ as ray $XL$.
3. Locate Vertex A: The third vertex $A$ of the required triangle $\triangle ABC$ is the point of intersection of the rays $XL$ and $YM$. Extend these rays until they intersect. Label the point of intersection as $A$.
4. Locate Vertex B: Vertex $B$ must lie on the segment $XY$. We need to locate $B$ such that the part $XB$ of the perimeter segment becomes equal to side $AB$. By the property of perpendicular bisectors, any point on the perpendicular bisector of $AX$ is equidistant from $A$ and $X$. Construct the perpendicular bisector of the line segment $AX$ (I3). Let this perpendicular bisector intersect the segment $XY$ at point $B$. By construction, $AB = XB$.
5. Locate Vertex C: Similarly, vertex $C$ must lie on the segment $XY$. We need to locate $C$ such that the part $YC$ of the perimeter segment becomes equal to side $AC$. Construct the perpendicular bisector of the line segment $AY$ (I3). Let this perpendicular bisector intersect the segment $XY$ at point $C$. By construction, $AC = YC$.
6. Complete Triangle: The vertices $A, B, C$ are now determined. The segment $BC$ is part of the original segment $XY$. Join point $A$ to point $B$ and point $A$ to point $C$.

The resulting figure $\triangle ABC$ is the required triangle. The side $BC$ is the segment of $XY$ between points $B$ and $C$.

Justification:

By construction, the segment $XY$ has length $P$. Vertex $A$ is formed by the intersection of rays $XL$ and $YM$, where $\angle YXL = \angle B/2$ and $\angle XYM = \angle C/2$.

Point $B$ lies on $XY$ and on the perpendicular bisector of $AX$. By the property of the perpendicular bisector, $AB = XB$.

Point $C$ lies on $XY$ and on the perpendicular bisector of $AY$. By the property of the perpendicular bisector, $AC = YC$.

Now consider the segment $XY$. It is composed of three parts: $XB$, $BC$, and $CY$.

Substitute $XB=AB$ and $YC=AC$ into equation (i): Since $XY$ was constructed with length $P$, this means $AB + BC + AC = P$, which is the given perimeter. The perimeter of $\triangle ABC$ is correct.

Now, let's examine the angles of $\triangle ABC$. Consider $\triangle ABX$. Since $AB = XB$, $\triangle ABX$ is an isosceles triangle. The angles opposite the equal sides are equal:

The exterior angle of $\triangle ABX$ at vertex $B$, which is $\angle ABC$, is equal to the sum of the two opposite interior angles ($\angle XAB$ and $\angle AXB$). From (iii) and (iv): The angle $\angle AXB$ is the same as $\angle YXL$, which was constructed as $\angle B/2$. Substitute (vi) into (v): Thus, $\angle ABC$ is equal to the given $\angle B$.

Similarly, consider $\triangle ACY$. Since $AC = YC$, $\triangle ACY$ is an isosceles triangle.

The exterior angle of $\triangle ACY$ at vertex $C$, which is $\angle ACB$, is equal to the sum of the two opposite interior angles ($\angle YAC$ and $\angle AYC$). From (viii) and (ix): The angle $\angle AYC$ is the same as $\angle XYM$, which was constructed as $\angle C/2$. Substitute (xi) into (x): Thus, $\angle ACB$ is equal to the given $\angle C$.

The triangle $ABC$ constructed has the given base angles $\angle B$ and $\angle C$ and the given perimeter $P$. This completes the justification.

Example

Answer: