Given a square $ABCD$, where $CE$ is parallel to $BD$, and $E$ belongs to $AD$. a) Prove that $\vec{AC} + \vec{BD} = \vec{AD} + \vec{BC}$. b) Prove that $\vec{AB} + \vec{BC} + \vec{CD} = \vec{AB} + \vec{CE}$.

1. Problem Description

Given a square $ABCD$, where $CE$ is parallel to $BD$, and $E$ belongs to $AD$.

a) Prove that $\vec{AC} + \vec{BD} = \vec{AD} + \vec{BC}$.

b) Prove that $\vec{AB} + \vec{BC} + \vec{CD} = \vec{AB} + \vec{CE}$.

2. Solution Steps

a) We need to prove $\vec{AC} + \vec{BD} = \vec{AD} + \vec{BC}$.

Since $ABCD$ is a square, $\vec{AC} = \vec{AB} + \vec{BC}$ and $\vec{BD} = \vec{BC} + \vec{CD}$.

Then $\vec{AC} + \vec{BD} = \vec{AB} + \vec{BC} + \vec{BC} + \vec{CD}$.

Since $ABCD$ is a square, $\vec{CD} = -\vec{AB}$.

Therefore, $\vec{AC} + \vec{BD} = \vec{AB} + 2\vec{BC} - \vec{AB} = 2\vec{BC}$.

Also, since $ABCD$ is a square, $\vec{AD} = \vec{BC}$.

So, $\vec{AD} + \vec{BC} = \vec{BC} + \vec{BC} = 2\vec{BC}$.

Therefore, $\vec{AC} + \vec{BD} = \vec{AD} + \vec{BC}$.

b) We need to prove $\vec{AB} + \vec{BC} + \vec{CD} = \vec{AB} + \vec{CE}$.

So, we need to prove that $\vec{BC} + \vec{CD} = \vec{CE}$.

$\vec{BC} + \vec{CD} = \vec{BD}$.

We are given that $CE$ is parallel to $BD$, which means $\vec{CE} = k\vec{BD}$ for some scalar $k$.

We are given that $E$ is on $AD$. Since $ABCD$ is a square, $AD$ is parallel to $BC$. Also, $CE$ is parallel to $BD$.

In $\triangle ABD$, $\angle DAB = 90^{\circ}$ and $\angle ABD = 45^{\circ}$.

In $\triangle BCE$, since $CE$ is parallel to $BD$, $\angle BCE = \angle CBD = 45^{\circ}$.

Since $E \in AD$, $\angle BCE = 45^{\circ}$. Then we can say that $CE = BD$.

Therefore, $\vec{BC} + \vec{CD} = \vec{BD} = \vec{CE}$.

3. Final Answer

a) $\vec{AC} + \vec{BD} = \vec{AD} + \vec{BC}$

b) $\vec{AB} + \vec{BC} + \vec{CD} = \vec{AB} + \vec{CE}$