# 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 $$C_1D_1 = A_1 D_1′ = A_1 B_1 + B_1 D_1’ = A_1 B_1 + B_1 D_1 = \ell_1 + \ell_2$$