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 O is point of intersection of CD and BE , prove that AO is the bisector of line segment DE
Answers
Solution:
I have drawn the diagram of this question below.
Just writing the explanation to reach at the answer:
Since points B,C,D and E are Concyclic.
Sum of opposite angles of 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.
Gives, ∠1 +∠4= 180°
∠2+∠3=180°
But these are interior angles on the same side of transversal B D and EC.So if sum of interior angles on the same side of transversal is 180°, then lines are parallel.
∵ DE ║ BC
∠4=∠5
∠3=∠6
when lines are parallel , corresponding angles are equal.
But ,∠3=∠4
Which gives, ∠5=∠6
∴ AD=A E→→If opposite angles in a triangle are equal , side opposite to them are equal.
Draw , AM ⊥ DE
In ΔAMD and ΔA ME
AD=AE →→Proved above
∠AMD=∠AME→→each being 90°.
AM is common.
ΔAMD ≅ ΔA ME→→[R HS congruency rule]
→DM = ME→→[CPCT]
Join ,AO,which will pass through M, as ∠AME +∠OME=180°,showing points A,M and O are collinear.
Which shows, Segment AO is the bisector of line segment DE because, DM = ME.
Answer:
ans. is above in that attachment
