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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. You may use sequential hints.

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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