Question

ABCD is a cyclic qudilatral whise diagonal at point e. if /dbc is70° /bac is 30°. find bcd further AB=BC find ECD​

Answer 1

The region between a chord and either of its arcs is called a segment the circle.

Angles in the same segment of a circle are equal.

For chord CD,

We know, that Angles in same segment are equal.

∠CBD =∠CAD+∠CAD=70°

∠BAD =∠BAC +∠CAD =

30° + 70° = 100°

ZBCD+ZBAD=180°

(Opposite angles of a cyclic quadrilateral)

∠BCD + 100° = 180°

∠BCD =180° - 100°

∠BCD=80°

In △ABC

AB = BC (given)

∠BCA =∠CAB

(Angles opposite to equal sides of a triangle)

∠CAD=30°

also, ∠BCD = 80°

∠BCA +∠ACD =80°

30° + ∠ACD = 80°

∠ACD =50°

∠ECD =50°

Hence,

∠BCD = 80° & ∠ECD=

50⁰