Angle Bisector Theorem

Step 1 (compare areas using the shared altitude from $A$). Triangles $\triangle ABD$ and $\triangle ACD$ share the same altitude from vertex $A$ down to the line $BC$, so their areas are proportional to their bases on $BC$: $\dfrac{S(\triangle ABD)}{S(\triangle ACD)} = \dfrac{BD}{DC}$.

Step 2 (compare the same two areas using $AB$ and $AC$ as bases). Because $D$ lies on the angle bisector of $A$, the perpendicular distances from $D$ to the two sides of the angle are equal: $d(D, AB) = d(D, AC)$. Taking $AB$ and $AC$ as the bases of $\triangle ABD$ and $\triangle ACD$, those two equal perpendiculars are the corresponding altitudes, so the area ratio also equals the ratio of these bases: $\dfrac{S(\triangle ABD)}{S(\triangle ACD)} = \dfrac{AB}{AC}$.

Step 3 (equate the two expressions). Since both right-hand sides equal the same area ratio, equating Step 1 and Step 2 immediately gives $\dfrac{BD}{DC} = \dfrac{AB}{AC}$, completing the proof.