Given any ∆ABC, let us bisect the angles at B and C. The bisectors meet at the incentre I of the triangle.
Now extend the bisectors beyond I till they meet the opposite sides at E and F respectively, as shown.
Proposition: If AB = AC, then IE = IF.

Step-by-step solution

Idea: For the proposition, equal sides give equal base angles, so the half-angles at B and C match and △IBF ≅ △ICE. For the converse, IE = IF can also happen in triangles that are not isosceles: it happens whenever ∠A = 60°, because then ∠BIC = 120° and two congruence steps link IF and IE.

1. Converse: If IE = IF, then AB = AC.½ mark
2. Proposition. AB = AC ⇒ ∠ABC = ∠ACB (angles opposite equal sides). BI and CI bisect these equal angles, so all four half-angles are equal: ∠IBC = ∠ICB and ∠FBI = ∠ECI.½ mark
3. In △IBC, ∠IBC = ∠ICB, so IB = IC (sides opposite equal angles).½ mark
4. In △IBF and △ICE: ∠FBI = ∠ECI, IB = IC, and ∠BIF = ∠CIE (vertically opposite angles, since B, I, E and C, I, F lie on straight lines). So △IBF ≅ △ICE (ASA) and IF = IE (CPCT). The proposition is true.1 mark
5. Converse — false. Take △ABC with ∠A = 60°, ∠B = 90°, ∠C = 30°, for example AB = 4 cm and AC = 8 cm. Here AB ≠ AC.½ mark
6. ∠IBC = 45° and ∠ICB = 15°, so ∠BIC = 180° − 60° = 120°. Hence ∠BIF = ∠CIE = 180° − 120° = 60°.½ mark
7. Mark D on BC with BD = BF. △BFI ≅ △BDI (SAS: BF = BD, ∠FBI = ∠DBI, BI common), so IF = ID and ∠BID = ∠BIF = 60°.½ mark
8. Then ∠DIC = ∠BIC − ∠BID = 120° − 60° = 60° = ∠EIC. △DIC ≅ △EIC (ASA: ∠DIC = ∠EIC, IC common, ∠DCI = ∠ECI), so ID = IE.½ mark
9. So IE = ID = IF (≈ 1.52 cm here), although AB = 4 cm ≠ 8 cm = AC. This triangle is a counterexample: the converse is false. (The same steps work for any triangle with ∠A = 60°.)½ mark

Check: With B(0, 0), A(0, 4) and C(4√3, 0): I ≈ (1.46, 1.46), E ≈ (2.54, 2.54), F ≈ (0, 1.86). Both IE and IF come out as ≈ 1.516 cm, while AB = 4 cm and AC = 8 cm.

Answer to write in the exam

Common mistakes that cost marks

• Assuming the converse is true because the figure looks symmetric. The counterexample triangle is far from isosceles.
• Using SAS for △IBF and △ICE without first proving IB = IC.
• Calling ∠BIF and ∠CIE equal ‘by symmetry’ instead of giving the reason: they are vertically opposite angles.

How this can come in the exam

Try one yourself

In △PQR, PQ = PR and the bisectors of ∠Q and ∠R meet at I. Show that IQ = IR.

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