Congruence and constructing triangles

What is it about?

A triangle has six parts: three sides $a$, $b$, $c$ and three angles $\alpha$, $\beta$, $\gamma$. The side $a = [BC]$ lies opposite the vertex $A$, $b = [CA]$ opposite $B$ and $c = [AB]$ opposite $C$; the angle $\alpha$ is at $A$, $\beta$ at $B$, $\gamma$ at $C$. Two triangles that can be moved, turned or reflected so that they fit exactly onto each other are called congruent.

Often three suitable parts are enough to fix a triangle uniquely. The congruence criteria tell you which ones:

The condition for SSS is called the triangle inequality: $a + b > c$, $a + c > b$ and $b + c > a$. If, for example, $a + b < c$, the circles around $A$ and $B$ do not reach each other; if $a + b = c$, they only touch on the segment $[AB]$ – then there is no “real” triangle.

Careful with SSA: If the given angle lies opposite the shorter side (e.g. $\alpha$ with $a < c$), the distance $h$ from $B$ to the free arm of $\alpha$ decides:

• $a < h$: The circle around $B$ does not reach the arm – no solution.
• $a = h$: The circle touches the arm – exactly one solution, with a right angle at $C$.
• $h < a < c$: The circle cuts the arm twice – two different triangles $ABC_1$ and $ABC_2$.

If instead $a \ge c$ (SsA), the second intersection lies behind $A$ on the extension of the arm and gives no triangle – the solution is unique.

AAA is not a congruence criterion: three angles only fix the shape. Because $\alpha + \beta + \gamma = 180°$, the third angle is already determined by the other two. The triangles are similar but of different sizes.

How to proceed: First draw a sketch and highlight the given parts in colour. Then check which congruence criterion fits. You usually start from the given side: draw the base, mark angles with the set square, transfer lengths with the compasses. The construction steps record what you do, e.g. for SSS: “1. Draw $[AB]$ with $c = 6\,\text{cm}$. 2. Draw a circle around $A$ with $r = b$. 3. Draw a circle around $B$ with $r = a$. 4. The circles meet at $C$. 5. Join $C$ to $A$ and $B$.”

Try it

1. When the page loads, the construction runs through once. Press “Construct again” and then step through line by line with “Next step”: ruler, compasses and set square work, and the matching line appears in the construction steps. Where would the second intersection of the circles be – and why does it not give a new triangle?
2. 3–4–5 triangle: Which angle is special here?
3. Drag the point $C$ or $B$ on the finished triangle. Which parts change as well? Then shrink $a$ and $b$ until the circles no longer meet – the red dimension line shows the gap. Example
4. SAS: Turn the arm of $\alpha$ with the round grip. Is there a setting where no triangle appears?
5. ASA: Set $\alpha + \beta$ to exactly $180°$ and then to more. What happens to the arms?
6. SSA with two solutions: Compare $\gamma_1$ and $\gamma_2$ in the result card. Shrink $a$ to $3\,\text{cm}$ and $2\,\text{cm}$, then make it longer than $6\,\text{cm}$ (SsA). How many solutions are there each time?
7. AAA: Three triangles with the same angles. Are they congruent?

Tasks

Can you construct a triangle with $a = 2\,\text{cm}$, $b = 3\,\text{cm}$ and $c = 6\,\text{cm}$? Give a reason without drawing. And with $a = 3\,\text{cm}$, $b = 4\,\text{cm}$, $c = 7\,\text{cm}$?

Given $c = 6\,\text{cm}$, $b = 4.5\,\text{cm}$ and $\alpha = 50°$. Which congruence criterion fits? Construct the triangle, write down the construction steps and measure $a$, $\beta$ and $\gamma$.

Construct the triangle with $c = 6\,\text{cm}$, $\alpha = 50°$ and $\beta = 60°$. First calculate $\gamma$, then measure $a$ and $b$.

Given $c = 6\,\text{cm}$, $\alpha = 30°$ and $a = 4\,\text{cm}$. Construct. Why are there two triangles?

Task 5: Exactly one solution

Keep $c = 6\,\text{cm}$ and $\alpha = 30°$. For which lengths $a$ are there no, exactly one or two triangles? Hint: How far is $B$ from the free arm?

Task 6: Same angles – same triangle? (Challenge)

Paul claims: “If two triangles agree in all three angles, they are congruent.” Prove him wrong with $\alpha = 50°$ and $\beta = 60°$. Show

Ideas for the classroom

• Use: about grade 7, congruence and triangle constructions. On the projector the simulation can construct step by step while the class draws along in their exercise books; afterwards the lines of the construction steps can be compared together.
• Task mode: With “Measure the missing parts” switched off (meas=0) the construction starts on a blank sheet and measurements and verdict stay hidden. Students construct on their own first and then compare with “Next step”.
• Model: The construction always starts from the base $[AB]$, with $C$ above it; the intersection below gives a reflected, i.e. congruent, triangle. Lengths are set to $0.1\,\text{cm}$, angles to $1°$; measured parts are shown with “≈”.
• Common stumbling blocks:
  • The sketch is skipped or the parts are assigned wrongly (e.g. $a$ to the wrong side).
  • For SAS an angle is used that does not lie between the two sides – that is SSA.
  • For SSA only one of the two solutions is drawn.
  • “Three angles fix a triangle.” – They only fix the shape, not the size.
  • When marking $\beta$, the set square is placed on the wrong side or the wrong scale is read.