Geometry of Triangles: CBSE Class 9 Exercise 7.2

Key Theorems on Isosceles Triangles

An isosceles triangle is a triangle with at least two equal sides. In Exercise 7.2 of Class 9, we focus heavily on two fundamental theorems that govern these triangles:

1. Theorem 7.2 (Angle-Side Theorem): Angles opposite to equal sides of an isosceles triangle are equal. This means if $AB = AC$ in $\triangle ABC$, then $\angle B = \angle C$.
2. Theorem 7.3 (Converse Theorem): The sides opposite to equal angles of a triangle are equal. If $\angle B = \angle C$ in $\triangle ABC$, then $AB = AC$.

These theorems allow us to switch seamlessly between side equalities and angle equalities during complex multi-step proofs. To apply them effectively, you must always look for the given properties of the triangle and use construction techniques (like angle bisectors or medians) to create congruent triangles. Understanding these theorems is key to unlocking all proofs in class 9 maths geometry of triangles ex 7 2.

How to Prove Theorem 7.2 Step-by-Step

1. Identify the Givens and Construct the Bisector — Start with an isosceles triangle $ABC$ where $AB = AC$. Draw the bisector of $\angle A$ which intersects the base $BC$ at point $D$. This construction divides the main triangle into two smaller triangles: $\triangle ABD$ and $\triangle ACD$.
2. Apply Congruence Rules (SAS) — Compare $\triangle ABD$ and $\triangle ACD$. We have $AB = AC$ (Given), $\angle BAD = \angle CAD$ (by construction, since $AD$ is the angle bisector), and $AD = AD$ (Common side). Thus, $\triangle ABD \cong \triangle ACD$ by the SAS (Side-Angle-Side) congruence rule.
3. Use CPCT to Conclude the Proof — Since the triangles are congruent, their corresponding parts must be equal. Therefore, $\angle ABD = \angle ACD$ by CPCT (Corresponding Parts of Congruent Triangles). This means $\angle B = \angle C$, proving our theorem.

Proving Triangles: Avoiding Common Mistakes

Many students lose marks in Exercise 7.2 due to simple structural and logical errors. Keep these tips in mind:

• Don't Assume What You Need to Prove: In converse proofs, do not use $AB = AC$ as a given fact if the question specifically asks you to show the triangle is isosceles.
• Specify the Congruence Rule: Always write down the exact congruence rule used (e.g., SAS, ASA, SSS) in your proofs. Writing 'by congruence' is insufficient for CBSE examiners.
• State Construction Clearly: If you draw a perpendicular bisector, an angle bisector, or a median, write a explicit statement under 'Construction' before starting the proof.

Practice Questions with Solutions

• Q: In an isosceles triangle $ABC$, with $AB = AC$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Join $A$ to $O$. Show that $OB = OC$. A: Step 1: In $\triangle ABC$, we are given $AB = AC$. Since angles opposite to equal sides are equal, we have $\angle B = \angle C$. Step 2: Take half of both sides: $\frac{1}{2} \angle B = \frac{1}{2} \angle C$. Since $OB$ and $OC$ are angle bisectors, this means $\angle OBC = \angle OCB$. Step 3: In $\triangle OBC$, since $\angle OBC = \angle OCB$, the sides opposite to equal angles must be equal. Therefore, $OB = OC$. Final answer: Proven that $OB = OC$ because sides opposite to equal angles in $\triangle OBC$ are equal.
• Q: In $\triangle ABC$, $AD$ is the perpendicular bisector of $BC$. Show that $\triangle ABC$ is an isosceles triangle in which $AB = AC$. A: Step 1: Consider $\triangle ABD$ and $\triangle ACD$. Since $AD$ is a perpendicular bisector, we have $BD = CD$ and $\angle ADB = \angle ADC = 90^\circ$. Step 2: Compare the two triangles: - $BD = CD$ (given) - $\angle ADB = \angle ADC = 90^\circ$ (given) - $AD = AD$ (common side) Step 3: By SAS congruence criterion, $\triangle ABD \cong \triangle ACD$. Step 4: By CPCT (Corresponding Parts of Congruent Triangles), we get $AB = AC$. Final answer: Since $AB = AC$, $\triangle ABC$ is an isosceles triangle.
• Q: $ABC$ and $DBC$ are two isosceles triangles on the same base $BC$. Show that $\angle ABD = \angle ACD$. A: Step 1: Since $ABC$ is an isosceles triangle on base $BC$, we have $AB = AC$. Therefore, $\angle ABC = \angle ACB$ (angles opposite to equal sides are equal). Step 2: Similarly, since $DBC$ is an isosceles triangle on base $BC$, we have $DB = DC$. Therefore, $\angle DBC = \angle DCB$. Step 3: Add the two equations: $\angle ABC + \angle DBC = \angle ACB + \angle DCB$. Step 4: Simplifying the angles, we get $\angle ABD = \angle ACD$. Final answer: Proven that $\angle ABD = \angle ACD$ by adding the equal base angles of both isosceles triangles.
• Q: $\triangle ABC$ is an isosceles triangle in which $AB = AC$. Side $BA$ is produced to $D$ such that $AD = AB$. Show that $\angle BCD$ is a right angle. A: Step 1: In $\triangle ABC$, $AB = AC$. Thus, $\angle ACB = \angle ABC$ (let this be $x$). Step 2: We are given $AD = AB$. Since $AB = AC$, we have $AD = AC$. In $\triangle ACD$, since $AD = AC$, we have $\angle ADC = \angle ACD = y$. Step 3: Now look at the full triangle $\triangle BCD$. The sum of angles is $\angle B + \angle BCD + \angle D = 180^\circ$. Substitute the values: $x + (x + y) + y = 180^\circ$. Step 4: This simplifies to $2x + 2y = 180^\circ$, which means $2(x+y) = 180^\circ$, or $x + y = 90^\circ$. Since $\angle BCD = x + y$, we have $\angle BCD = 90^\circ$. Final answer: $\angle BCD$ is a right angle ($90^\circ$).

Frequently Asked Questions