a) Proof: Let $\triangle ABC$ be a triangle such that $\angle B \cong \angle C$. We need to prove that the sides opposite to these angles are congruent, i.e., $AC \cong AB$. Step 1: Draw the angle bisector of $\angle A$. Draw a line segment $AD$ from vertex $A$ to side $BC$ such that $AD$ bisects $\angle A$. This means $\angle BAD \cong \angle CAD$. Step 2: Consider the two triangles formed. Now we have two triangles: $\triangle ABD$ and $\triangle ACD$. Step 3: Apply the Angle-Angle-Side (AAS) congruence criterion. In $\triangle ABD$ and $\triangle ACD$: $\angle B \cong \angle C$ (Given) $\angle BAD \cong \angle CAD$ (By construction, $AD$ is the angle bisector of $\angle A$) $AD \cong AD$ (Common side to both triangles) Therefore, by the AAS congruence criterion, $\triangle ABD \cong \triangle ACD$. Step 4: Conclude based on CPCTC. Since the two triangles are congruent, their corresponding parts are congruent (CPCTC). Thus, the side $AB$ in $\triangle ABD$ corresponds to the side $AC$ in $\triangle ACD$. Hence, $AB \cong AC$. This proves that if two angles of a triangle are congruent, then the sides opposite to them are also congruent. b) Given the two polynomials: $P(x) = x^2 - 11x + 24$ $Q(x) = x^2 - 6x + 9$ And their HCF is $H(x) = (x-3)$. We need to find the LCM. Step 1: Factorize the first polynomial $P(x)$. We look for two numbers that multiply to 24 and add up to -11. These numbers are -3 and -8. $$ P(x) = x^2 - 11x + 24 = (x-3)(x-8) $$ Step 2: Factorize the second polynomial $Q(x)$. The polynomial $x^2 - 6x + 9$ is a perfect square trinomial, which can be factored as $(x-3)^2$. $$ Q(x) = x^2 - 6x + 9 = (x-3)(x-3) = (x-3)^2 $$ Step 3: Use the relationship between HCF, LCM, and the polynomials. The product of two polynomials is equal to the product of their HCF and LCM: $$ P(x) \times Q(x) = H(x) \times L(x) $$ Where $L(x)$ is the LCM. We can rearrange this to find $L(x)$: $$ L(x) = \frac{P(x) \times Q(x)}{H(x)} $$ Step 4: Substitute the factored forms and the given HCF into the formula. $$ L(x) = \frac{((x-3)(x-8)) \times ((x-3)^2)}{(x-3)} $$ Step 5: Simplify the expression to find the LCM. Cancel one $(x-3)$ term from the numerator and the denominator. $$ L(x) = (x-8)(x-3)^2 $$ We can expand this expression: $$ L(x) = (x-8)(x^2 - 6x + 9) $$ $$ L(x) = x(x^2 - 6x + 9) - 8(x^2 - 6x + 9) $$ $$ L(x) = x^3 - 6x^2 + 9x - 8x^2 + 48x - 72 $$ $$ L(x) = x^3 - 14x^2 + 57x - 72 $$ The LCM of the two polynomials is $\boxed{(x-8)(x-3)^2}$ or $\boxed{x^3 - 14x^2 + 57x - 72}$.