Rotation of triangle (B.Stat 2006, Problem 4 solution)

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 $E_1 D_1' = E_1 D_1 \implies \angle E_1 D_1' D_1 = \angle E_1 D_1 D_1'$

But $\angle E_1 D_1' B_1 = \angle E_1 D_1 B_1' = 90^o \implies \angle E_1 D_1' B_1 - \angle E_1 D_1' D_1= \angle E_1 D_1 B_1' - \angle E_1 D_1 D_1' \implies \angle B_1 D_1' D_1 = \angle B_1 D_1 D_1' \implies B_1 D_1 = B_1 D_1 '$

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

Rotation of triangle (B.Stat 2006, Problem 4 solution)

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