in the given figure ab and cd are two parallel chords of a circle.if bde and ace are straight line,intersecting at e,prove that aeb is isosceles triangle
Answers
ΔAEB is isosceles triangle ab and cd are two parallel chords of a circle.if bde and ace are straight line intersecting at e
Step-by-step explanation:
Given that
AB || CD
=> ∠ACD + ∠CAB = 180°
ABCD is a cyclic quadrilateral
=> ∠ACD + ∠ABD = 180° ( sum of opposite angles of cyclic Quadrilateral = 180°)
Equating both
∠ACD + ∠CAB = ∠ACD + ∠ABD
=> ∠CAB = ∠ABD
as C lies on AE & D lies on BE
=> ∠EAB = ∠ABE
=> AE = BE
=> ΔAEB is isosceles triangle
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Answer:
hello friends, question 16 exercise 12 C.
Step-by-step explanation:
given that
AB||CD
angleACD+angleCAB=180°
ABCD is a cyclic quadrilateral.
angleACD+angleABD=180°
angleACD+angleCAB=angleACD=angleABD
angleCAB=angleABD
as C lies on AE and D lies on BE.
angleEAB=angleABC
AE=BE
so, triangle AEB is isosceles triangle