Question

Given: ∠A≅∠C and BD bisects ∠ABC.
Prove: AC ⊥ BD.

Answer 1

Given triangle ABC with ∠A ≅ ∠C. Let D be the point on AC where BD meets AC and BD bisects ∠ABC (so ∠ABD = ∠DBC). Prove AC ⟂ BD.

Proof:

1. From ∠A ≅ ∠C, the sides opposite those equal angles are equal: AB = BC. (Isosceles triangle theorem.)

2. BD bisects ∠ABC, so ∠ABD = ∠DBC. (Given.)

3. In triangles ABD and CBD:
- AB = CB (from 1),
- ∠ABD = ∠DBC (from 2),
- BD = BD (common side).
Thus by SAS, ΔABD ≅ ΔCBD.

4. From the congruence, corresponding angles ∠ADB and ∠BDC are equal. But A, D, C are collinear, so ∠ADB and ∠BDC form a linear pair and hence are supplementary: ∠ADB + ∠BDC = 180°.

5. Two equal angles that sum to 180° must each be 90°. Therefore ∠ADB = ∠BDC = 90°, so BD ⟂ AC.

Hence AC is perpendicular to BD. □