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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 \(\mathbf{\ell_1, \ell_2} \).



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


Screen Shot 2015-05-08 at 3.35.28 PM

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

Screen Shot 2015-05-08 at 3.35.36 PM

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 \(C_1D_1 = A_1 D_1′ = A_1 B_1 + B_1 D_1’ = A_1 B_1 + B_1 D_1 = \ell_1 + \ell_2 \)

May 8, 2015

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