M is the mid-point of a line segment AB; AXB and MYB are equilateral triangles on opposite sides of AB; XY cuts AB at Z. Prove that : AZ = 2ZB.

In △XZB and △MZY,

∠XZB = ∠MZY (Vertically opposite angles are equal)

∠XBZ = ∠YMZ (Each = 60°)

△XZB ~ △MZY [By AA postulate]

In similar triangles,

The ratios between the lengths of corresponding sides are equal.

∴ $\dfrac{ZB}{MZ} = \dfrac{BX}{MY}$

As sides of equilateral triangle are equal so,

BX = AB

and

MY = MB

∴ $\dfrac{BX}{MY} = \dfrac{AB}{MB}$

∴ $\dfrac{ZB}{MZ} = \dfrac{BX}{MY} = \dfrac{AB}{MB}$ ……(1)

As, M is the mid-point of AB.

∴ MB = $\dfrac{1}{2}AB$

From (1),

$\Rightarrow \dfrac{ZB}{MZ} = \dfrac{AB}{MB} \\[1em] \Rightarrow \dfrac{ZB}{AZ - AM} = \dfrac{AB}{\dfrac{1}{2}AB} \\[1em] \Rightarrow \dfrac{ZB}{AZ - AM} = 2 \\[1em] \Rightarrow ZB = 2(AZ - AM) \\[1em] \Rightarrow ZB = 2AZ - 2AM \\[1em] \Rightarrow ZB = 2AZ - AB \space [\because \text{M is mid-point of AB}] \\[1em] \Rightarrow ZB = AZ + AZ - AB \\[1em] \Rightarrow ZB = AZ - (AB - AZ) \\[1em] \Rightarrow ZB = AZ - ZB \\[1em] \Rightarrow 2ZB = AZ.$

Hence, proved that AZ = 2ZB.