Solve AB^2+CD^2=BD^2+AC^2

$$A B ^ { 2 } + C D ^ { 2 } = B D ^ { 2 } + A C ^ { 2 }$$

Solve for A (complex solution)

$$\left\{\begin{matrix}A=\frac{D^{2}}{B+C}\text{, }&B\neq -C\\A\in \mathrm{C}\text{, }&B=C\text{ or }\left(D=0\text{ and }B=-C\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract AC^{2} from both sides.

$$AB^{2}+CD^{2}-AC^{2}=BD^{2}$$

Subtract CD^{2} from both sides.

$$AB^{2}-AC^{2}=BD^{2}-CD^{2}$$

Combine all terms containing A.

$$\left(B^{2}-C^{2}\right)A=BD^{2}-CD^{2}$$

Divide both sides by B^{2}-C^{2}.

$$\frac{\left(B^{2}-C^{2}\right)A}{B^{2}-C^{2}}=\frac{\left(B-C\right)D^{2}}{B^{2}-C^{2}}$$

Dividing by B^{2}-C^{2} undoes the multiplication by B^{2}-C^{2}.

$$A=\frac{\left(B-C\right)D^{2}}{B^{2}-C^{2}}$$

Divide \left(B-C\right)D^{2} by B^{2}-C^{2}.

$$A=\frac{D^{2}}{B+C}$$

Solve for A

$$\left\{\begin{matrix}A=\frac{D^{2}}{B+C}\text{, }&B\neq -C\\A\in \mathrm{R}\text{, }&B=C\text{ or }\left(D=0\text{ and }B=-C\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract AC^{2} from both sides.

$$AB^{2}+CD^{2}-AC^{2}=BD^{2}$$

Subtract CD^{2} from both sides.

$$AB^{2}-AC^{2}=BD^{2}-CD^{2}$$

Combine all terms containing A.

$$\left(B^{2}-C^{2}\right)A=BD^{2}-CD^{2}$$

Divide both sides by B^{2}-C^{2}.

$$\frac{\left(B^{2}-C^{2}\right)A}{B^{2}-C^{2}}=\frac{\left(B-C\right)D^{2}}{B^{2}-C^{2}}$$

Dividing by B^{2}-C^{2} undoes the multiplication by B^{2}-C^{2}.

$$A=\frac{\left(B-C\right)D^{2}}{B^{2}-C^{2}}$$

Divide \left(B-C\right)D^{2} by B^{2}-C^{2}.

$$A=\frac{D^{2}}{B+C}$$

Solve for B (complex solution)

$$\left\{\begin{matrix}\\B=C\text{, }&\text{unconditionally}\\B=\frac{D^{2}}{A}-C\text{, }&A\neq 0\\B\in \mathrm{C}\text{, }&D=0\text{ and }A=0\end{matrix}\right.$$

Solve for B

$$\left\{\begin{matrix}\\B=C\text{, }&\text{unconditionally}\\B=\frac{D^{2}}{A}-C\text{, }&A\neq 0\\B\in \mathrm{R}\text{, }&D=0\text{ and }A=0\end{matrix}\right.$$