I.S.I. and C.M.I. Entrance

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} .

B.Stat Entrance 2006 Geometry Problem


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


B.Stat Entrance 2006 Geometry Problem

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

By Dr. Ashani Dasgupta

Ph.D. in Mathematics, University of Wisconsin, Milwaukee, United States.

Research Interest: Geometric Group Theory, Relatively Hyperbolic Groups.

Founder, Cheenta

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