1. ## triangle

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2. Originally Posted by helloying
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Similarity:

$\angle BOA = \angle COA$

$90^o - \angle BAO = \angle OBA = \angle OAC$

$\angle OAB = \angle OCA$

3. Originally Posted by helloying
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As angles BAC and BDA are right angles angle BAD is congrunet to to angle ACD (amoung other reasons because triangles BAD and BAC are both right and have a common angle other than the right angle)

Hence triangles ABD and CAD are both right triangles and another pair of congruent andgles and so are similar.

Part (b) you do by forming ratios of corresponding sides of two of the similar triangles.

Part (c) yuo use the result of part (b) and Pythogoras' theorem to find AD, then the area is (1/2)AD.BD and BD is given.

RonL

4. Originally Posted by helloying
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Let AO = x. Since the triangles are similar you have the following equality of side ratios: $\frac{4}{x} = \frac{x}{9}$ ....

It should be simple to get the area now.