Triangle Squared

Construct point $A'$ such that $YAZA'$ is a parallelogram.

First, we will show that $A Y A' \cong ABC$ by $SAS$ . Observe that $A' Y = AZ = AC$ and $YA = AB$ . Also, since $\angle A'YA = 180^{\circ} - \angle YAZ = \angle BAC$ , hence the result follows. We also get that $\angle AA'Y = \angle CBA$ and $AA' = BC$ .

Second, we will show that $A', T, A, D$ are collinear. We already know that $T, A, D$ are collinear, and need to account for $A'$ . Observe that

$\begin{array} { l l l } \angle A'AY & = \angle CBA & \text{ similar triangles } \\ & = 90 - \angle BAD & \text{ angle chasing }\\ & = \angle TAY & \text{ angles on a line } \\ \end{array}$

Since $AY$ makes the same angle as $AA'$ and $AT$ , hence we conclude that $A', A, T$ are collinear.

Third, since $T$ lies on both $AA'$ and $YZ$ , hence it is the intersection of the diagonals of the parallelogram. This implies that $T$ is the midpoint of $A'A$ so $AT = \frac{ AA'}{2} = \frac{BC}{2} = \boxed{2}$

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Let 0, b, c be the complex numbers corresponding to the vertices A,B and C. Then Y is (-bi) and Z is (ci). By section formula, complex number corresponding to T is k(ci)+(1-k)(-bi) for some scalar k in [0,1].Now since, AT is perpendicular to BC, their complex slopes sum to 0.This on simplifying will give k=1/2.Hence, complex number corresponding to T is (i/2)(c-b).Hence AT=(1/2)|c-b|=(1/2)BC=2.