a) Prove that if two triangles \( \triangle ABC \) and \( \triangle A'B'C' \) are similar, then the lengths of their sides are proportional in order: \[ \frac{|AB|}{|A'B'|} = \frac{|BC|}{|B'C'|} = \frac{|CA|}{|C'A'|} \] Diagram: Given: - \( \triangle ABC \) and \( \triangle A'B'C' \) are similar - Leaving Cert Mathematics - Question 6 - 2021

The lengths of the sides of the triangles are proportional in order: [ \frac{|AB|}{|A'B'|} = \frac{|BC|}{|B'C'|} = \frac{|CA|}{|C'A'|} ]

Mark ( B'' ) on ( AB ) such that ( |AB''| = |A'B'| ). Mark ( C'' ) on ( AC ) such that ( |AC''| = |A'C'| ). Join ( B''C'' ).

1. ( \triangle A'B'C' ) is congruent to ( \triangle ABC ) (Reason: SAS).
2. Therefore, ( B''C'' \parallel BC ) due to corresponding angles: ( \angle A'B'C' = \angle ABC ).
3. This implies that ( \frac{|AB|}{|A'B'|} = \frac{|AC|}{|A'C'|} ) and similarly for the other sides.

Lines ( PA, HK, ) and ( BR ) are parallel.

|AP| \times |QB| = |AP| \times |HB|.

1. ( |HQ| = |HB| ) alternate segments.
2. ( \angle QHB = \angle QAP ) are equal because they are alternate interior angles (PA || HK).
3. Since triangles are similar, we can state that ( |AP| \times |QB| = |AP| \times |HB| ).