I2.3 Given that $\triangle A B C$ is an isosceles triangle, $A B=A C=\sqrt{2}$, and $D_{1}, D_{2}, \ldots, D_{Q}$ are $Q$ points on $B C$. Let $m_{i}=A D_{i}^{2}+B D_{i} \times D_{i} C$. If $m_{1}+m_{2}+m_{3}+\ldots+m_{Q}=R$, find the value of $R$.
As shown in the figure, $A B=A C=\sqrt{2}$
$$
B D=x, C D=y, A D=t, \angle A D C=\theta
$$

Apply cosine formula on $\triangle A B D$ and $\triangle A C D$
$$
\begin{array}{l}
\cos \theta=\frac{t^{2}+y^{2}-2}{2 t y} \\
\cos \left(180^{\circ}-\theta\right)=\frac{t^{2}+x^{2}-2}{2 t x}
\end{array}
$$
since $\cos \left(180^{\circ}-\theta\right)=-\cos \theta$
Add these equations and multiply by $2 t x y$ :
$$
\begin{array}{l}
x\left(t^{2}+y^{2}-2\right)+y\left(t^{2}+x^{2}-2\right)=0 \\
(x+y) t^{2}+(x+y) x y-2(x+y)=0 \\
(x+y)\left(t^{2}+x y-2\right)=0 \\
t^{2}+x y-2=0 \\
A D^{2}+B D \cdot D C=2 \\
R=m_{1}+m_{2}+m_{3}+m_{4}=2+2+2+2=8
\end{array}
$$

Method 2
Let $B D=x, C D=y, A D=t, \angle A B C=\alpha=\angle A C D$,
$$
\angle B A D=\theta, \angle C A D=\varphi \text {. }
$$

Rotate $A D$ anticlockwise about $A$ to $A E$ so that
$$
\begin{array}{l}
\angle D A E=\angle B A C . \\
\angle C A E=\angle D A E-\varphi=\angle B A D=\theta
\end{array}
$$

By the property of rotation, $A E=A D=t$.
$$
\begin{array}{l}
\triangle C A E \cong \triangle B A D \\
C E=B D=x \\
\angle A C E=\angle A B D=\alpha \\
\angle D A E+\angle D C E=\theta+\varphi+2 \alpha=180^{\circ} \\
\Rightarrow 2 \alpha=180^{\circ}-(\theta+\varphi) \cdots \cdots \cdots\left(^{*}\right)
\end{array}
$$
(corr. sides, $\cong \Delta \mathrm{s}$ )
(corr. $\angle \mathrm{s}, \cong \Delta \mathrm{s}$ )
( $\angle \mathrm{s}$ sum of $\Delta$ )

The area of $A D C E=S_{\triangle A D E}+S_{\triangle C E}=$ the area of $\triangle A B C$
$$
\begin{array}{l}
\frac{1}{2} t^{2} \sin (\varphi+\theta)+\frac{1}{2} x y \sin 2 \alpha=\frac{1}{2} \sqrt{2}^{2} \sin (\varphi+\theta) \\
t^{2} \sin (\theta+\varphi)+x y \sin \left[180^{\circ}-(\theta+\varphi)\right]=2 \sin (\theta+\varphi) \text { by }(*) \\
\because \sin \left[180^{\circ}-(\theta+\varphi)\right]=\sin (\theta+\varphi) \therefore t^{2}+x y=2 \\
A D^{2}+B D \cdot D C=2 \\
R=m_{1}+m_{2}+m_{3}+m_{4}=2+2+2+2=8
\end{array}
$$