The symbol on a signboard needs to be recreated on another board. One way is to trace the outline, but this is difficult for big symbols.
To recreate a figure exactly, specific measurements are needed. For a symbol with corner points A, B, and C, the arm lengths AB and BC alone are not sufficient.
If $AB = 4$ cm and $BC = 8$ cm, several different symbols can be constructed. To get an exact replica, the measure of $\angle ABC$ along with the two arm lengths AB and BC, fix the shape and size of the figure.
Figures that are exact copies of each other or have the same shape and size are said to be congruent. Congruent figures can be superimposed exactly, one over the other. When checking for congruence, a figure can be rotated or flipped before superimposing.
Meera and Rabia need to make a cardboard cutout identical to a triangular frame. The frame is too big to trace.
The girls measure the sides of the triangle to be $40$ cm, $60$ cm, and $80$ cm. Meera states that the angles are not required; the side lengths are sufficient to create a congruent triangle.
If the sidelengths were $4$ cm, $6$ cm, $8$ cm, Rabia suggests drawing a line segment (e.g., $6$ cm), then drawing circles from its endpoints with radii $4$ cm and $8$ cm. These circles intersect at two points, forming two triangles: $\triangle ABE$ and $\triangle ABF$.
These two triangles, $\triangle ABE$ and $\triangle ABF$, are congruent. This can be verified by tracing, cutting out, or observing that AB acts as a line of symmetry. This shows that all triangles with the same sidelengths are congruent.
Thus, if two triangles have the same sidelengths, then they are congruent. This is called the SSS (Side Side Side) condition for congruence.
When two triangles are congruent, they can be superimposed so that equal sides overlap. For $\triangle ABC$ and $\triangle XYZ$ to be congruent, the corresponding vertices must align:
This relation is written as $\triangle ABC \cong \triangle XYZ$. This notation implies that the first vertex in $\triangle ABC$ corresponds to the first in $\triangle XYZ$, and similarly for the second and third vertices. For example, it is incorrect to write $\triangle ACB \cong \triangle XYZ$, but $\triangle ACB \cong \triangle XZY$ is correct.
Consider $\triangle ABD$ and $\triangle CDB$ in a rectangle ABCD. Since ABCD is a rectangle, $AB = CD$ and $AD = CB$. The remaining side BD is common. Thus, the SSS condition is satisfied, confirming the congruence of the two triangles.
If Meera and Rabia measure only the three angles of the triangular frame (e.g., $30^\circ$, $70^\circ$, and $80^\circ$), they cannot recreate an exact copy. Many triangles can be drawn with these measurements that are not congruent, as they can have the same shape but different sizes.
Consider $\triangle ABC$ and $\triangle XYZ$ such that $AB = XY = 6$ cm, $AC = XZ = 5$ cm, and $\angle A = \angle X = 30^\circ$. These measurements correspond to two sides and the included angle. When two sides and the included angle of two triangles are equal, the two triangles are congruent. This is referred to as the SAS (Side Angle Side) condition for congruence.
What if two sides and a non-included angle are equal? For $\triangle ABC$ and $\triangle XYZ$ with $AB = XY = 6$ cm, $AC = XZ = 4$ cm, and $\angle B = \angle Y = 30^\circ$, there can exist non-congruent triangles. This is called the SSA (Side Side Angle) condition, which does not guarantee congruence.
Construction example: Draw base PQ ($6$ cm). Draw a line $l$ from P at $30^\circ$ to PQ. Draw an arc from Q with radius $4$ cm. The arc intersects line $l$ at two different points, R and S, forming two triangles ($\triangle PQR$ and $\triangle PQS$) that satisfy the given measurements but are not congruent.
Consider $\triangle ABC$ and $\triangle XYZ$ with $BC = YZ = 5$ cm, $\angle B = \angle Y = 50^\circ$, and $\angle C = \angle Z = 30^\circ$. This is the case of two angles and the included side. This construction makes it clear that all triangles with these measurements must be congruent. This is referred to as the ASA (Angle Side Angle) condition for congruence.
In a figure where Point O is the midpoint of AD and BC, we have $AO = OD$ and $BO = OC$. Also, $\angle AOB = \angle DOC$ (vertically opposite angles). By SAS condition, $\triangle AOB \cong \triangle DOC$. Thus, $AB = DC$ (corresponding sides).
Consider $\triangle ABC$ and $\triangle XYZ$ such that $\angle A = \angle X = 35^\circ$, $\angle C = \angle Z = 75^\circ$, and $BC = YZ = 4$ cm. This involves two angles and a non-included side.
Using the angle sum property of a triangle ($180^\circ$), we find $\angle B = 180^\circ - (35^\circ + 75^\circ) = 70^\circ$. Similarly, $\angle Y = 70^\circ$. Now, we have $\angle B = \angle Y$, $BC = YZ$, and $\angle C = \angle Z$. This satisfies the ASA condition, so the triangles are congruent.
This condition is referred to as the AAS (Angle Angle Side) condition. AAS condition guarantees congruence.
Consider right-angled triangles $\triangle ABC$ and $\triangle XYZ$ such that $BC = YZ = 4$ cm, $\angle B = \angle Y = 90^\circ$, and $AC = XZ = 5$ cm. This is a special case of SSA. The side opposite the right angle is called the hypotenuse.
When a side and a hypotenuse of a right-angled triangle are equal to a side and the hypotenuse of another right-angled triangle, the triangles are congruent. This is called the RHS (Right Hypotenuse Side) condition for congruence.
The RHS condition is one more condition that guarantees congruence.
From the discussions, two triangles are congruent if any of the following conditions are satisfied:
Note that the SSA condition does not always guarantee congruence.
Congruence is a powerful tool for studying properties of geometric figures.
Consider an isosceles triangle $\triangle ABC$ with $AB = AC$ and $\angle A = 80^\circ$. Construct the altitude from A to BC. This forms two right-angled triangles, $\triangle ADB$ and $\triangle ADC$.
We have: $AB = AC$ (given), $\angle ADB = \angle ADC = 90^\circ$ (from construction), and AD is a common side. Thus, by RHS condition, $\triangle ADB \cong \triangle ADC$. This shows that $\angle B = \angle C$, as they are corresponding parts of congruent triangles. Therefore, the angles opposite to equal sides are equal.
For an equilateral triangle, all sides have equal lengths. Since $AB = AC$, then $\angle B = \angle C$. Since $AB = BC$, then $\angle A = \angle C$. Therefore, all three angles are equal. As the sum of angles in a triangle is $180^\circ$, each angle in an equilateral triangle is $180^\circ / 3 = 60^\circ$.
Congruent triangles are seen in various constructions and designs in real life, such as the Louvre Museum, Egyptian Pyramids, dome designs, and Rangoli designs.