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# 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

Challenges and Thrills - Pre - College Mathematics

## Try with Hints

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 ......................................................

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

Apply Pythagoras Theorem ,

$AB ^2 = AC^2 + BC ^2$

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

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

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

$y = \frac {12}{5}$

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 .

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