In the diagram below, \( \triangle ABC \) and \( \triangle PQR \) are given with \( \hat{A} = \hat{P}, \hat{B} = \hat{Q} \) and \( \hat{C} = \hat{R} - NSC Mathematics - Question 10 - 2016 - Paper 2

Given that ( \triangle ADE ) is similar to ( \triangle PQR ), it follows that the corresponding sides and angles are equal. Therefore, by corresponding angles,

If ( \angle ADE = \angle PQR ) and ( \angle ABE = \angle RQP ), then

( DE \parallel BC ) by the Converse of the Corresponding Angles Postulate.

Since we have established that ( DE \parallel BC ), we can apply the Basic Proportionality Theorem (or Thales' theorem). This theorem states that if a line is drawn parallel to one side of a triangle, then it divides the other two sides proportionally:

[ \frac{AB}{PQ} = \frac{AC}{PR} ]

The angle ( OP ) is perpendicular to ( PS ) because by the property of circles, a radius drawn to the point of tangency is perpendicular to the tangent. In this context, ( OP ) is the radius, and since S lies on the circumference, ( OP \perp PS ) holds.

Since ( P ) is the midpoint of ( RS ) and the line segments ( OP ) and ( VS ) are transversals to the parallel lines, we have:

( \frac{AR}{RO} = \frac{VS}{AR} ) Thus, by the Basic Proportionality Theorem, the triangles are similar, leading to: [ \triangle AROP \parallel \triangle ARVS ]

By properties of parallel lines and similar triangles, since transversals intersect the parallel lines, we can set up the ratio and angles: [ AR ext{ is the same in both triangles. Since angles are preserved,} ARVS \parallel ARST ]

By the Extended Law of Proportions in right triangles and basic circle properties, we have shown that: [ R_{T}S = V_{T}S = 90^{\circ} ext{ which establishes that } ST \text{ is a tangent, leading to } ST^{2} = VT \times TR. ]