Triangles – Congruence, Properties & Inequalities

Congruency in geometry means two figures or objects are identical in shape and size or one is the mirror image of the other. They can be transformed into each other through rigid motions: translation, rotation, and reflection, without resizing. This means one figure can be repositioned and flipped to coincide exactly with the other. Congruent figures have equal corresponding sides, angles, diagonals, perimeters, and areas (e.g., congruent triangles have three pairs of equal sides and angles). Congruency is denoted by the symbol "≅". Unlike similarity, congruent figures are the same size as well as shape. This concept applies to various geometric shapes such as line segments, angles, triangles, and circles, where if the measurements match, the shapes are congruent

Two triangles are said to be congruent if they are exactly equal in both shape and size. This means that all the corresponding sides and angles of the two triangles are equal. Congruent triangles can be perfectly superimposed on each other without any gaps or overlaps.

Note that in congruent triangles corresponding parts are equal and we write in short ‘CPCT’ for corresponding parts of congruent triangles.

The criteria for congruence of triangles are five essential rules used to prove that two triangles are exactly equal in shape and size. These criteria are:

1. SSS (Side-Side-Side) Criterion:
   If all three sides of one triangle are equal to the corresponding three sides of another triangle, then the two triangles are congruent.
2. SAS (Side-Angle-Side) Criterion:
   If two sides and the included angle (the angle between those two sides) of one triangle are equal to the corresponding two sides and included angle of another triangle, the triangles are congruent.
3. ASA (Angle-Side-Angle) Criterion:
   If two angles and the included side (the side between those two angles) of one triangle are equal to the corresponding two angles and included side of another triangle, the triangles are congruent.
4. AAS (Angle-Angle-Side) Criterion:
   If two angles and a non-included side of one triangle are equal to the corresponding two angles and side of another triangle, then the triangles are congruent.
5. RHS (Right angle-Hypotenuse-Side) Criterion:
   In right-angled triangles, if the hypotenuse and one other side of one triangle are equal to the hypotenuse and one corresponding side of another triangle, the triangles are congruent.

In Fig. 7.8, OA = OB and OD = OC. Show that (i) \(\Delta AOD \cong \Delta BOC\) and (ii) \(AD \parallel BC\)

Given $$\begin{aligned}AO&=BO\\ DO&=CO\\ \angle OB&=\angle AOD\end{aligned}$$ (i) To prove $$\triangle AOD\cong \triangle BOC$$ Proof

In \(\triangle AOD\) and \(\triangle BOC\)

$$\begin{aligned}AO&=BO\\ DO&=CO\\ \angle COB&=\angle AOD\end{aligned}$$ $$\triangle AOD\cong \triangle BOC$$ (By SAS rule of Congruency)
Hence Proved. $$ AD\parallel BC$$ (ii) To prove $$\angle OCB=\angle ODA$$ (By CPCT) (Alternate Angles for the lines AD and CB) $$ \therefore AD\parallel BC$$ Hence Proved

\(AB\) is a line segment and line \(l\) is its perpendicular bisector. If a point P lies on \(l\), show that \(P\) is equidistant from \(A\) and \(B\).

Given:

\(AB\) is a line segment and \(l\) is \(\perp \;AB\)

To Prove
\[AP=BP\]
Prrof

In \(\triangle APC\) and \(\triangle BPC\)
(\(C\) is a midpoint as \(l\) is perpedicular bisector of \(AB\)) : Given

$$\therefore AC=CB$$ $$\begin{aligned}\angle ACP&=\angle BCP\\ &=90^{\circ }\end{aligned}$$ PC = PC (common) $$\triangle APC\cong \triangle BPC \\\text{ (By SAS Rule of Congruency)}$$ \[AP = BP\text{ (By  CPCT)}\]
Hence Proved

Line-segment \(AB\) is parallel to another line-segment \(CD\). \(O\) is the mid-point of \(AD\) (see Fig. 7.15). Show that (i) \(\triangle AOB \cong \triangle DOC\) (ii) \(O\) is also the mid-point of \(BC\).

Given:

\(AB\parallel CD\) and \(AO = OD\) (\(O\) is midpoint of \(AD)\)

To prove $$\triangle AOB\cong \triangle DOC$$ Proof
$$AO = OD \;\text{ (Given)}$$ $$\angle DCO=\angle ABO\\\text{(Alternate Angle)}$$ $$\angle AOB=\angle DOC\\\text{(Vertically opposite angles)}$$ $$\Delta AOB\cong \triangle DOC\\\text{(By AAS rule of Congruency)}$$ Hence Proved. $$CO = OB \\\text{(by CPCT)}$$

\(O\) is also mid-point of \(BC\)

 Proved.

A triangle in which two sides are equal is called an isosceles triangle.

Angles opposite to equal sides of an isosceles triangle are equal.

Proof:

Given \(\triangle ABC\) is an isocles triangle \(AB = AC\)

To Prove
$$\angle ABC=\angle ACD$$ Construction:

Draw a angle bisector of \(\angle A\), and let \(D\) be the point of intersection of this angle bisector

In \(\triangle BAD\) and \(\triangle CAD\)

$$AB=AC \text{ (Given)}$$ $$AD=AD\text{ (Common)}$$ $$\angle BAD=\angle CAD\text{ (By Construction)}$$ $$\triangle BAD\cong \triangle CAD\text{ (By SAS Rule)}$$ $$\angle ABC=\angle ACD$$ Hence Proved.

The sides opposite to equal angles of a triangle are equal.

In \(\triangle ABC\), the bisector \(AD\) of \(\angle A\) is perpendicular to side \(BC\) (see Fig. 7.27). Show that \(AB = AC\) and \(\triangle ABC\) is isosceles.

In \(\triangle ABD\) and \(\triangle ACD\)

$$\angle BAD=\angle CAD\text{ (Given)}$$ \[AD = AD \text{ (common)}\] $$\angle ADB=\angle ADC=90^{\circ }\text{ (Given)}$$ $$\triangle ABD\cong \triangle ACD\\\text{(ASA Rule of Congruency)}$$ $$\therefore AB=AC\text{ (By CPCT)}$$

\(\triangle ABC\) is an isosceles triangle

E and F are respectively the mid-points of equal sides AB and AC of ∆ ABC (see Fig. 7.28). Show that BF = CE.

Solution:

In \(\triangle ABF\) and \(\triangle ACE\)

\[\begin{align} AB &= AC\text { (Given)}\\ AF &= AE \text{ (Half of AB and AG)}\\ \angle A&=\angle A\text{ (Common)}\end{align}\] \[\triangle ABF\cong \triangle ACE\\\text{(SAS Rule of Congruency)}\] therefore \[BF = CE \text{ (By CPCT)}\]

AB is a line-segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (see Fig. 7.37). Show that the line PQ is the perpendicular bisector of AB.

Solution:

\(AP = BP\) (Given)
\(AQ = BQ\) (Given)

To show that
\(PQ\) is a ⟂ bisector of \( AB\)

We need to prove that
$$AC = BC\\ \angle ACP=\angle BCP=90^{\circ }$$ Proof:

In \(\triangle PAQ\) and \(\triangle PBQ\)

\[\begin{aligned}AP &= BP\text{ (Given)} \\AQ &= BQ \text{ (Given)} \\PQ &= PQ \text{ (Common)}\end{aligned}\] $$\therefore \Delta PAQ\cong \Delta PBQ\\\text{ (SSS Rule)}$$ $$\Rightarrow \angle APC=\angle BPC\text{ (By CPCT)}\tag{1}$$ \[\small\begin{aligned}\text{In } \triangle PAC \text{ and} \triangle PBC\\ AP &= BP \text{ (Given)}\\ PC &= PC \text{ (Common)}\\ \angle APC&=\angle BPC\text{ (From Eqn (I))}\end{aligned}\] $$\Delta PAC\cong \Delta PBC\text{ (SAS Rule)}$$ $$\therefore \angle PCA=\angle PCB\text{ (By CPCT)}$$ $$AC=BC\text{ (By CPCT)}$$ $$\begin{aligned}\angle ACP+\angle BCP&=180^{\circ}\text{ (Linear Pair)}\\ 2\angle ACP&=180^{\circ}\\ \angle ACP&=90^{\circ}\end{aligned}$$ Proved.