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Equal side and angle

Source: Jim Noble on Twitter

In the diagram shown, the lengths

A D = C D

and the angles

A B D = C B D

.

Prove that the lengths

A B = B C

.

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By the sine rule in BCD:

\frac{C D}{\sin (C B D)} = \frac{B D}{\sin (B C D)}

By the sine rule in BAD:

\frac{A D}{\sin (A B D)} = \frac{B D}{\sin (B A D)}

A B D = C B D

and

A D = C D

, so:

\sin (B A D) = \frac{B D \sin (A B D)}{A D} = \frac{B D \sin (C B D)}{C D} = \sin (B C D)

Using the sine rule in ABC:

\frac{A B}{\sin (B C D)} = \frac{B C}{\sin (B A D)}

The two sines are equal and so:

A B = B C

Tags: geometry, 2d shapes, triangles, trigonometry

If you enjoyed this puzzle, check out Sunday Afternoon Maths XXXVIII,
puzzles about 2d shapes, or a random puzzle.

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