# Application of Pythagoras Theorem | SMO, 2010 | Problem 22

Try this problem from the Singapore Mathematics Olympiad, SMO, 2010 based on the application of the Pythagoras Theorem.

## Application of Pythagoras Theorem- (SMO Test)

The figure below shows a circle with diameter AB. C ad D are points on the circle on the same side of AB such that BD bisects \(\angle {CBA}\). The chords AC and BD intersect at E. It is given that AE = 169 cm and EC = 119 cm. If ED = x cm, find the value of x.

- 65
- 55
- 56
- 60

**Key Concepts**

Circle

Pythagoras Theorem

2D - Geometry** **

## Check the Answer

But try the problem first...

Answer: 65

Singapore Mathematical Olympiad

Challenges and Thrills - Pre - College Mathematics

## Try with Hints

First hint

If you get stuck in this problem this is the first hint we can start with :

As BE intersect \(\angle {CBA}\) we have \(\frac {BC}{BA} = \frac {EC}{EA} = \frac {119}{169}\)

Thus we can let BC = 119 y and BA = 169 y .

Since \(\angle {BCA} = 90 ^\circ\).

Then try to do the rest of the problem ......................................................

Second Hint

If we want to continue from the last hint we have :

Apply Pythagoras Theorem ,

\(AB ^2 = AC^2 + BC ^2\)

\((169y)^2 = (169 + 119)^2 + (119y)^2\)

\(y^2 (169-119)(169+119) = (169+119)^2\)

\(y^2 = \frac {169+119}{169-119} = \frac {144}{25}\)

\(y = \frac {12}{5}\)

Final Step

In the last hint:

Hence , from triangle BCE , we have BE = \(\sqrt{119^2 + (119y)^2} = 119 \times \frac {13}{5}\)

Finally , note that \(\triangle {ADE}\) and \(\triangle {BCE}\) are similar , so we have

ED = \(\frac {AE \times CE}{BE} = \frac {169 \times 119}{119 \times \frac {13}{5}} = 65 \) cm .

## Other useful links

- https://www.cheenta.com/functional-equations-problem-smo-2012-problem-33/
- https://www.youtube.com/watch?v=pLAMlNUOdTs