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Explore the Back-Story**Problem:**

In the figure below, $E$ is the midpoint of the arc $ABEC$ and the segment $ED$ is perpendicular to the chord $BC$ at $D$. If the length of the chord $AB$ is $\mathbf{\ell_1} $, and that of the segment $BD$ is $\mathbf{\ell_2} $, determine the length of $DC$ in terms of $\mathbf{\ell_1, \ell_2} $.

**Discussion:**

**Teacher:** Here is a clue: rotate ($ \Delta EDC$ ) about point $E$ such that $EC$ falls along $EA$. Can you draw the diagram after rotation?

**Student:**

Obviously $EC$ will fit into $EA$ as $E$ is the midpoint of larger arc $AC$. Suppose $D$ falls on $D'$. Then $ED'A$ is the rotated form of $EDC$.

**Teacher: **Can you show that $D'A$ and $BA$ are the same line?

**Student:** Okay I can try. If we can show that $\angle D'AE = \angle BAE $ , then we have shown that $D'A$ and $BA$ is the same line. Now $\angle D'AE = \angle DCE = \angle BCE$ (due to rotation) . But $\angle BCE = \angle BAE $ as they are the angle subtended by the same segment $BE$.

So we have $\angle D'AE = \angle BAE $. Therefore $D'A$ falls on $BA$.

**Teacher:** Let's revise the diagram then.

Now can you finish the problem?

**Student:** Since $ \Delta E_1 D_1' A_1 = \Delta E_1 D_1 C_1 $ due to rotation, we have $E_1 D_1' = E_1 D_1 \Rightarrow \angle E_1 D_1' D_1 = \angle E_1 D_1 D_1' $

But $\angle E_1 D_1' B_1 = \angle E_1 D_1 B_1' = 90^{\circ} \\ \Rightarrow \angle E_1 D_1' B_1 - \angle E_1 D_1' D_1 \\ = \angle E_1 D_1 B_1' - \angle E_1 D_1 D_1' \Rightarrow \angle B_1 D_1' D_1 \\ = \angle B_1 D_1 D_1' \Rightarrow B_1 D_1 \\ = B_1 D_1 ' $

Thus $ C_1D_1 = A_1 D_1' \\ = A_1 B_1 + B_1 D_1' \\= A_1 B_1 + B_1 D_1 \\= \ell_1 + \ell_2 $

**Problem:**

In the figure below, $E$ is the midpoint of the arc $ABEC$ and the segment $ED$ is perpendicular to the chord $BC$ at $D$. If the length of the chord $AB$ is $\mathbf{\ell_1} $, and that of the segment $BD$ is $\mathbf{\ell_2} $, determine the length of $DC$ in terms of $\mathbf{\ell_1, \ell_2} $.

**Discussion:**

**Teacher:** Here is a clue: rotate ($ \Delta EDC$ ) about point $E$ such that $EC$ falls along $EA$. Can you draw the diagram after rotation?

**Student:**

Obviously $EC$ will fit into $EA$ as $E$ is the midpoint of larger arc $AC$. Suppose $D$ falls on $D'$. Then $ED'A$ is the rotated form of $EDC$.

**Teacher: **Can you show that $D'A$ and $BA$ are the same line?

**Student:** Okay I can try. If we can show that $\angle D'AE = \angle BAE $ , then we have shown that $D'A$ and $BA$ is the same line. Now $\angle D'AE = \angle DCE = \angle BCE$ (due to rotation) . But $\angle BCE = \angle BAE $ as they are the angle subtended by the same segment $BE$.

So we have $\angle D'AE = \angle BAE $. Therefore $D'A$ falls on $BA$.

**Teacher:** Let's revise the diagram then.

Now can you finish the problem?

**Student:** Since $ \Delta E_1 D_1' A_1 = \Delta E_1 D_1 C_1 $ due to rotation, we have $E_1 D_1' = E_1 D_1 \Rightarrow \angle E_1 D_1' D_1 = \angle E_1 D_1 D_1' $

But $\angle E_1 D_1' B_1 = \angle E_1 D_1 B_1' = 90^{\circ} \\ \Rightarrow \angle E_1 D_1' B_1 - \angle E_1 D_1' D_1 \\ = \angle E_1 D_1 B_1' - \angle E_1 D_1 D_1' \Rightarrow \angle B_1 D_1' D_1 \\ = \angle B_1 D_1 D_1' \Rightarrow B_1 D_1 \\ = B_1 D_1 ' $

Thus $ C_1D_1 = A_1 D_1' \\ = A_1 B_1 + B_1 D_1' \\= A_1 B_1 + B_1 D_1 \\= \ell_1 + \ell_2 $

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