Q.If $P_1, P_2, P_3$ are altitudes drawn from vertices A, B, C to the opposite sides of a triangle respectively, then show that:

Area of a Triangle

There is no single "best" area formula — which one to reach for depends entirely on what data the problem hands you. This chapter collects four equivalent formulas, each optimal for a different combination of known sides/angles:

1. SAS form: $\Delta=\frac12bc\sin A$ (and cyclically) — use this the moment you know two sides and the included angle.
2. Heron's formula: $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$, $s=\frac{a+b+c}2$ — use this for SSS, when no angle is given at all.
3. $\Delta=\dfrac{a^2\sin B\sin C}{2\sin A}$ — use this for ASA/AAS, when one side and (effectively) all three angles are known.
4. $\Delta=2R^2\sin A\sin B\sin C$ — use this when the circumradius $R$ is known alongside the three angles.

How they connect

All four are really the same formula in disguise. Formula (1) is the base case, proved directly from "half base times height." Formula (2) follows by substituting the half-angle expressions for $\sin\frac A2,\cos\frac A2$ into $\sin A=2\sin\frac A2\cos\frac A2$ and then into (1). Formula (3) follows by eliminating $c$ from (1) using the sine rule ($c=\frac{a\sin C}{\sin A}$). Formula (4) follows by eliminating both $b$ and $c$ from (1) using $b=2R\sin B,c=2R\sin C$. Recognising this family resemblance means you only truly need to remember (1) and derive the rest on demand.

A common trap …