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A, B, C, and D are four points on a circle as shown - Junior Cycle Mathematics - Question 10 - 2014

Question 10

A, B, C, and D are four points on a circle as shown. [AD] bisects ∠BAC. P is the point of intersection of AD and BC. (i) Show that △ADB and △APC are similar. (ii) ... show full transcript

Worked Solution & Example Answer:A, B, C, and D are four points on a circle as shown - Junior Cycle Mathematics - Question 10 - 2014

Step 1

Show that △ADB and △APC are similar.

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Answer

To demonstrate that triangles △ADB and △APC are similar, we start by recognizing that [AD] bisects ∠BAC. This implies that:

$|∠BAD| = |∠PAC|$

Additionally, since angles ADB and APC share angle APB, we can state that:

$|∠ADB| + |∠DBA| + |∠PAB| = 180°$

From this, we conclude that:

$|∠DBA| = |∠CAP|$

Consequently, the angles of triangles ADB and APC are equal, leading us to conclude that the triangles are similar. Thus, we establish:

$△ADB \sim △APC$

Step 2

Show that |AC| × |BD| = |AD| × |PC|.

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Answer

With the knowledge that △ADB and △APC are similar, we can apply the property that corresponding sides of similar triangles are in proportion. Therefore, we have:

$\frac{|AC|}{|AD|} = \frac{|PC|}{|BD|}$

Cross-multiplying the equation gives us:

$|AC| × |BD| = |AD| × |PC|$

This confirms the required relationship between the lengths of the segments.

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