Question

D and e are respectively the points on equal sides ab and ac of an isosceles triangle abc such that b, c, e and d are concyclic. If m' is the point of intersection of cd and be, prove that am is the bisector of line segment de

Answer 1

Since points B,C,D and E are Con cyclic

(Sum of opposite angles)

cyclic quadrilateral is 180°

So , ∠1 + ∠3= 180°

and , ∠2 + ∠4= 180°

Also , ΔABC is Isosceles Triangle

AB= AC ----- (Given)

∠3=∠4 -----  (when opposite sides are equal angle opposite to them are equal)

Given , ∠1 +∠4= 180°

∠2+∠3=180°

But , they are interior angles on the same side of transversal BD and EC , So, If the sum of interior angles on same side of transversal is 180°, than lines are parallel

∵ DE ║ BC

∠4=∠5

∠3=∠6

lines are parallel , corresponding angles are equal.

But ,∠3=∠4

Which given, ∠5=∠6

∴ AD=A E-----   (If opposite angles in a triangle are equal , side opposite to them are equal)

Draw , AM ⊥ DE

In ΔA MD and ΔA ME

AD=A E

∠A MD=∠A ME (each  90°)

AM =AM (common)

ΔA MD ≅ ΔA ME (R.H.S)

→DM = ME ( by C.P.C.T)

Join A O, pass through M, as ∠AME +∠O ME=180°,show the  points A,M and O are collinear.

Which showing , Segment A O is the bisector of line segment DE because, D M = ME.