A circle touches the side BC of a $\triangle ABC$ at a point P and touches AB and AC when produced at Q and R respectively. Show that $AQ = \frac{1}{2}$ (Perimeter of $\triangle ABC$).
Generated by claude-sonnet-4-6 · 2026-06-15 10:27 · grounding rag
Model Answer
To Prove: $AQ = \dfrac{1}{2}$ (Perimeter of $\triangle ABC$)
Proof:
Since tangents drawn from an external point to a circle are equal in length (Theorem 10.2):
- From external point A: $AQ = AR$ ... (i)
- From external point B: $BQ = BP$ ... (ii)
- From external point C: $CP = CR$ ... (iii)
Now, Perimeter of $\triangle ABC$:
$$= AB + BC + CA$$
$$= AB + (BP + PC) + CA$$
$$= AB + BQ + CR + CA \quad \text{[using (ii) and (iii)]}$$
$$= (AB + BQ) + (CR + CA)$$
$$= AQ + AR \quad \text{[since } AQ = AB + BQ \text{ and } AR = AC + CR\text{]}$$
$$= AQ + AQ \quad \text{[using (i)]}$$
$$= 2 \cdot AQ$$
$$\therefore AQ = \frac{1}{2} \text{ (Perimeter of } \triangle ABC\text{)}$$
Hence proved.
Source: Chapter 10, Section 10.3 (Theorem 10.2)
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Explanation
- The key property used is Theorem 10.2: tangents from an external point are equal. This applies to all three vertices A, B, C as external points.
- Note that Q and R lie on the produced sides AB and AC, so $AQ = AB + BQ$ and $AR = AC + CR$ — students often miss this sign.
- Examiners award marks for: correctly stating the equal tangent pairs (1 mark each pair), the algebraic substitution leading to Perimeter = 2AQ (2 marks), and the conclusion (1 mark).
- Write the tangent-length equalities explicitly — do not skip steps.