Given a triangle ABC, we are asked to solve three problems related to a point M in the plane. 1) Prove that there exists a unique point D such that $\vec{DA} + \vec{DB} - \vec{DC} = \vec{0}$. 2) Prove that for any point M in the plane, $\vec{MA} + \vec{MB} - \vec{MC} = \vec{MD}$. 3) Determine the set of points M in the plane such that: a) $\vec{MA} + \vec{MB} - \vec{MC}$ is collinear with $\vec{AM} - \vec{BM}$. b) $||\vec{MA} + \vec{MB} - \vec{MC}|| = ||\vec{MB}||$. c) $||\vec{MA} + \vec{MB} - \vec{MC}|| = 1$.

1. Problem Description

Given a triangle ABC, we are asked to solve three problems related to a point M in the plane.

1) Prove that there exists a unique point D such that $\vec{DA} + \vec{DB} - \vec{DC} = \vec{0}$.

2) Prove that for any point M in the plane, $\vec{MA} + \vec{MB} - \vec{MC} = \vec{MD}$.

3) Determine the set of points M in the plane such that:

a) $\vec{MA} + \vec{MB} - \vec{MC}$ is collinear with $\vec{AM} - \vec{BM}$.

b) $||\vec{MA} + \vec{MB} - \vec{MC}|| = ||\vec{MB}||$.

c) $||\vec{MA} + \vec{MB} - \vec{MC}|| = 1$.

2. Solution Steps

1) Existence and uniqueness of D such that $\vec{DA} + \vec{DB} - \vec{DC} = \vec{0}$.

$\vec{DA} + \vec{DB} - \vec{DC} = \vec{0}$

$\vec{DA} + \vec{DB} + \vec{CD} = \vec{0}$

$\vec{CD} = \vec{AD} - \vec{BD}$

$\vec{DC} = \vec{BD} - \vec{AD}$

$\vec{AD} = \vec{AB} + \vec{BD}$

$\vec{BD} - \vec{AD} = \vec{BD} - (\vec{AB} + \vec{BD}) = -\vec{AB}$

Therefore, $\vec{DC} = -\vec{AB}$, which means $\vec{CD} = \vec{AB}$.

This means that vector $\vec{CD}$ is equal to vector $\vec{AB}$, thus the quadrilateral ABDC is a parallelogram. Since A, B, and C are fixed points, D is uniquely defined as the fourth vertex of parallelogram ABDC.

The intersection of the diagonals AC and BD is the midpoint.

2) For any point M in the plane, prove that $\vec{MA} + \vec{MB} - \vec{MC} = \vec{MD}$.

We know that $\vec{DA} + \vec{DB} - \vec{DC} = \vec{0}$.

Using Chasles' relation,

$\vec{MA} = \vec{MD} + \vec{DA}$

$\vec{MB} = \vec{MD} + \vec{DB}$

$\vec{MC} = \vec{MD} + \vec{DC}$

Therefore,

$\vec{MA} + \vec{MB} - \vec{MC} = (\vec{MD} + \vec{DA}) + (\vec{MD} + \vec{DB}) - (\vec{MD} + \vec{DC})$

$\vec{MA} + \vec{MB} - \vec{MC} = \vec{MD} + \vec{DA} + \vec{MD} + \vec{DB} - \vec{MD} - \vec{DC}$

$\vec{MA} + \vec{MB} - \vec{MC} = \vec{MD} + \vec{DA} + \vec{DB} - \vec{DC}$

$\vec{MA} + \vec{MB} - \vec{MC} = \vec{MD} + (\vec{DA} + \vec{DB} - \vec{DC})$

Since $\vec{DA} + \vec{DB} - \vec{DC} = \vec{0}$,

$\vec{MA} + \vec{MB} - \vec{MC} = \vec{MD}$

3)

a) $\vec{MA} + \vec{MB} - \vec{MC}$ is collinear with $\vec{AM} - \vec{BM}$.

Since $\vec{MA} + \vec{MB} - \vec{MC} = \vec{MD}$, we have $\vec{MD}$ is collinear with $\vec{AM} - \vec{BM}$.

$\vec{AM} - \vec{BM} = (\vec{AD} + \vec{DM}) - (\vec{BD} + \vec{DM}) = \vec{AD} - \vec{BD} = \vec{AD} + \vec{DB}$

Since $\vec{DA} + \vec{DB} - \vec{DC} = \vec{0}$, we have $\vec{DB} = -\vec{DA} + \vec{DC}$

So, $\vec{AD} - \vec{BD} = \vec{AD} - (-\vec{DA} + \vec{DC}) = \vec{AD} + \vec{DA} - \vec{DC} = 2\vec{DA} - \vec{DC}$

We have $\vec{MD}$ is collinear with $2\vec{DA} - \vec{DC}$. This means that M, D, and a point E such that $\vec{DE} = 2\vec{DA} - \vec{DC}$ are collinear. Since D, A and C are fixed, the point E is also fixed.

Therefore, M lies on the line (DE).

b) $||\vec{MA} + \vec{MB} - \vec{MC}|| = ||\vec{MB}||$

Since $\vec{MA} + \vec{MB} - \vec{MC} = \vec{MD}$, we have $||\vec{MD}|| = ||\vec{MB}||$.

This means that the distance from M to D is the same as the distance from M to B. Therefore, M lies on the perpendicular bisector of segment BD.

c) $||\vec{MA} + \vec{MB} - \vec{MC}|| = 1$

Since $\vec{MA} + \vec{MB} - \vec{MC} = \vec{MD}$, we have $||\vec{MD}|| = 1$.

This means that the distance from M to D is

1. Therefore, M lies on the circle centered at D with radius

1.

3. Final Answer

1) D is uniquely defined as the fourth vertex of parallelogram ABDC.

2) $\vec{MA} + \vec{MB} - \vec{MC} = \vec{MD}$

3)

a) M lies on the line (DE), where E is a fixed point such that $\vec{DE} = 2\vec{DA} - \vec{DC}$.

b) M lies on the perpendicular bisector of segment BD.

c) M lies on the circle centered at D with radius

1.