Question

Triangle ABC has incentre I and the incircle touches BC,CA at D,E respectively.Let BI be produced to meet DE at G.Show that AG perpendicular to BG.

Answer 1

AEGI is a cyclic quadrilateral

∠EGI = ∠ GBD + ∠GDB (External angle of triangle BGD )

= ∠B/2 + (∠BDI+∠IDG)

= ∠B/2 + (90°+∠IDE)  (∠BDI = 90° as angle with the tangent and chord joining with center is 90°)

= ∠B/2+(90°+∠ICE)  (∠IDE=∠ICE angles is same segment )

= ∠B/2+90°+∠C/2 (∠C/2=∠ICE because in center is made by the angle bisectors point of contact)

∠B+∠C/2+90°

= 180°−∠A/2+90°

= 90°−∠A/2+90°

Thus, ∠EGI=180°−∠A/2

Hence , AEGI is a cyclic quadrilateral with diameter as AI.

Therefore, ∠AGI=∠AGB =90° and AG ⊥ BG.