Isosceles Triangles

218. The Bisector's Balance · Why AF always splits the vertex angle

AF always bisects both the vertex angle ∠BAC and the base BC, however tall the triangle is.

In an isosceles triangle the line from the apex to the base can be the angle bisector, the median, and the altitude at once. Here AF bisects ∠BAC, splitting the vertex angle into two equal halves, because it creates two congruent triangles (SAS) whose matching angles are equal.

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Selina ICSE: Isosceles Triangles

What this lesson covers

Try to break it

Drag A up or down. The triangle stays isosceles (AB = AC by symmetry), so the angle bisectors from B and C are mirror images and always meet on the centre line at F. AF splits ∠A in half. Try to find a height where the bisectors miss the centre line — impossible.

How you build it

Construct an isosceles triangle and its medians.

Point tool: mark point B, one end of the base.
Point tool: mark point C — BC is the base.
Segment tool: join B to C.
Bisector tool: click B then C. This draws the perpendicular bisector of BC — the triangle's line of symmetry.
Point tool: drop the apex A anywhere on the perpendicular bisector. Because A sits on it, AB = AC, so triangle ABC is isosceles.
Segment tool: join A to B.
Segment tool: join A to C.
Midpoint tool: click A then B to mark P, the midpoint of side AB.
Segment tool: join C to P — the median to side AB.
Midpoint tool: click A then C to mark Q, the midpoint of side AC.
Segment tool: join B to Q — the median to side AC. All three medians meet at one balance point on the axis, and the apex line (the perpendicular bisector) bisects both the base BC and the vertex angle ∠BAC.

The proof, step by step

Prove that AF bisects ∠BAC.

In ΔABC, AB = AC (Given) ⇒ ∠ABC = ∠ACB (Base angles of isosceles triangle are equal).
BF and CF bisect ∠ABC and ∠ACB respectively ⇒ ∠FBC = ∠FCB = ½∠ABC.
In ΔBFC, ∠FBC = ∠FCB ⇒ FB = FC (Sides opposite to equal angles are equal).
In ΔAFB and ΔAFC: AB = AC, FB = FC, AF = AF (Common) ⇒ ΔAFB ≅ ΔAFC (SSS Congruence).
∠FAB = ∠FAC (C.P.C.T.C.) ⇒ AF bisects ∠BAC. Hence Proved.

Worked example

In ΔABC, AB = AC. The angle bisectors of ∠B and ∠C intersect at F. The line AF is:

In an isosceles triangle, the vertex angle bisector coincides with the altitude and median to the base. Since F lies on the bisector of ∠A, AF serves all three roles simultaneously.

The altitude from A only
The median to BC only
The angle bisector of ∠A only
The altitude, median, and angle bisector from A — correct