Q.$AB$ is a diameter of a circle and $C$ is any point on the circle. Show that the area of $\triangle ABC$ is maximum when it is isosceles.

Let the circle's radius be $r$, so $AB = 2r$ is fixed. By Thales' theorem, $\angle ACB = 90^\circ$, so with $\angle CAB = \theta$:

Step 1: Area as a function of $\theta$: $\Delta(\theta) = \dfrac{1}{2}AC\cdot BC = r^2\sin2\theta$.

Step 2: Differentiate: $\dfrac{d\Delta}{d\theta} = 2r^2\cos2\theta = 0 \Rightarrow \theta = \dfrac{\pi}{4}$ (since $\theta \in (0,\frac{\pi}{2})$). …

Setting up

Let the circle have radius $r$, so the diameter $AB = 2r$ is a fixed length (the circle itself doesn't change — only the position of $C$ on it does). Let $C$ be any point on the circle, and let $\angle CAB = \theta$, where $\theta$ can range over $\left(0, \dfrac{\pi}{2}\right)$ as $C$ moves around the semicircle.

Since $AB$ is a diameter and $C$ lies on the circle, the angle in a semicircle is a right angle (Thales' theorem):

So $\triangle ABC$ is always right-angled at $C$, with the fixed segment $AB$ as its hypotenuse.

Step 1 — Express the legs in terms of $\theta$

In this right triangle, with hypotenuse $AB = 2r$ and angle $\theta$ at $A$:

This turns the problem into a single-variable optimization in $\theta$ — no partial derivatives or multiple constraints are needed.

Step 2 — Write the area as a function of $\theta$

Step 3 — Differentiate and find the critical point

Setting $\dfrac{d\Delta}{d\theta} = 0$:

Step 4 — Confirm it's a maximum

$\frac{d^2\Delta}{d\theta^2} = -4r^2\sin2\theta.$ …

Method: Constrained Optimization via Lagrange Multipliers

Use this whenever you must maximize or minimize some quantity built from two related lengths, while a point is forced to stay on a fixed curve (a circle, in most geometry problems of this type).

Steps

Step 1: Choose the two variables and write the objective function.

Pick the two quantities whose product or combination you're actually optimizing (here, the two legs of a right triangle inscribed in a circle). Call them $x$ and $y$, and write the quantity to be maximized as $f(x,y)$ — for an area built from two perpendicular legs, this is typically $f(x,y)=xy$ (a constant factor of $\tfrac12$ doesn't change where the maximum occurs, so it can be dropped for convenience).

Step 2: Write the constraint as $g(x,y)=0$.

A fixed diameter forces the point on the circle to obey Pythagoras: whatever is fixed (the diameter, or diagonal, or hypotenuse) gives a relation like $x^2+y^2=d^2$, written as

Step 3: Set up the Lagrange condition $\nabla f=\lambda\nabla g$.

Compute both gradients and equate component-wise:

This is the geometric statement that at the optimum, moving along the constraint curve can no longer increase $f$ — the two gradients must point in the same (or opposite) direction. …

Common Mistakes

Mistake 1: Treating the diameter $AB$ as if it can also vary

Why it's wrong: The problem fixes $AB$ as the diameter of a given circle — only point $C$ moves along the circle. A student who lets both the diameter and $C$'s position vary is optimising the wrong number of variables and will not get a clean isosceles-triangle result. Correct approach: fix $AB=2R$ (or $d$) as a constant, and treat only $C=(x,y)$ on $x^2+y^2=R^2$ as the variable point.

Mistake 2: Not using Thales' theorem to fix the right angle at $C$

Why it's wrong: Since $AB$ is a diameter, $\angle ACB=90^\circ$ for every position of $C$ on the circle — this is what lets the area be written simply as $\tfrac12\cdot AC\cdot BC$ under the constraint $AC^2+BC^2=AB^2$. Missing this forces a much messier (and error-prone) coordinate area formula. Correct approach: recognise the right angle at $C$ first, then set up $f=AC\cdot BC$ subject to $AC^2+BC^2=d^2$.

Mistake 3: Finding the critical point but never checking it is a maximum …