🔑 Important Formulas, Rules & Key Points

• Congruent Figures: Two geometric figures are congruent if they have the same shape and size. In triangles, congruence is represented by the symbol ≅.
• CPCT (Corresponding Parts of Congruent Triangles): If two triangles are proven congruent, their corresponding angles and sides are equal.
• SAS Congruence Rule (Side-Angle-Side): Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.
• ASA Congruence Rule (Angle-Side-Angle): Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.
• AAS Congruence Rule (Angle-Angle-Side): Two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal.

Question 1: In quadrilateral $ABCD$, $AC = AD$ and $AB$ bisects $\angle A$. Show that $\triangle ABC \cong \triangle ABD$. What can you say about $BC$ and $BD$?

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Given:
1. $AC = AD$
2. Line segment $AB$ bisects $\angle A$, which means $\angle CAB = \angle DAB$.

To Prove:
$\triangle ABC \cong \triangle ABD$ and determine the relation between $BC$ and $BD$.

Proof:
Consider $\triangle ABC$ and $\triangle ABD$:
1. $AC = AD$ (Given)
2. $\angle CAB = \angle DAB$ (Given, as $AB$ is the bisector of $\angle A$)
3. $AB = AB$ (Common side)

Therefore, by SAS Congruence Rule:
$$\triangle ABC \cong \triangle ABD$$

Now, by CPCT (Corresponding Parts of Congruent Triangles):
$$BC = BD$$

Conclusion: $BC$ and $BD$ are of equal length.

Question 2: $ABCD$ is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$. Prove that:
(i) $\triangle ABD \cong \triangle BAC$
(ii) $BD = AC$
(iii) $\angle ABD = \angle BAC$

---

Given:
$AD = BC$ and $\angle DAB = \angle CBA$.

Proof:
(i) To prove $\triangle ABD \cong \triangle BAC$:
In $\triangle ABD$ and $\triangle BAC$:
1. $AD = BC$ (Given)
2. $\angle DAB = \angle CBA$ (Given)
3. $AB = BA$ (Common side)

By SAS Congruence Criterion:
$$\triangle ABD \cong \triangle BAC$$

(ii) To prove $BD = AC$:
Since $\triangle ABD \cong \triangle BAC$ (proved above), we have:
$$BD = AC \quad [\text{by CPCT}]$$

(iii) To prove $\angle ABD = \angle BAC$:
Similarly, since the triangles are congruent:
$$\angle ABD = \angle BAC \quad [\text{by CPCT}]$$

Question 3: $AD$ and $BC$ are equal perpendiculars to a line segment $AB$. Show that $CD$ bisects $AB$.

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Given:
$AD = BC$ and both are perpendicular to $AB$ (i.e., $\angle OAD = \angle OBC = 90^\circ$). Let $O$ be the intersection point of $CD$ and $AB$.

To Prove:
$CD$ bisects $AB$, which means we need to show $OA = OB$.

Proof:
Consider $\triangle OAD$ and $\triangle OBC$:
1. $\angle OAD = \angle OBC = 90^\circ$ (Given)
2. $\angle AOD = \angle BOC$ (Vertically opposite angles)
3. $AD = BC$ (Given)

Therefore, by AAS Congruence Rule:
$$\triangle OAD \cong \triangle OBC$$

By CPCT:
$$OA = OB$$

Since $OA = OB$, the point $O$ is the midpoint of $AB$. Hence, $CD$ bisects $AB$.

🧠 Aptitude Corner (NTSE / Olympiad Focus)

Concept Booster Question:
In two triangles $PQR$ and $XYZ$, if $PQ = XY$, $QR = YZ$, and $\angle PQR = \angle XYZ = 60^\circ$. What will be the nature of the relationship between their perimeters?

Solution:
By applying the SAS congruence criterion, we find $\triangle PQR \cong \triangle XYZ$. Since congruent triangles have exactly identical shapes and sizes, all corresponding sides are equal ($PQ=XY, QR=YZ,$ and $PR=XZ$ via CPCT). Therefore, the sum of their sides must be equal: $$\text{Perimeter of } \triangle PQR = \text{Perimeter of } \triangle XYZ$$

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