eral whose diagonals intersect at a point E. If 4 DBCABCD is a cycliBA

1.	eral whose diagonals intersect at a point E. If 4 DBCABCD is a cycliBAC is 30°, find ZBCD. Further, ifAB -BC, find 4 ECD.70°6.lic quadrilateral are diameters of the circle through the vertices of
Answer» The region between a chord andeither of its arcs is called a segment the circle. Angles in the same segment of a circle areequal. For chord CD, We know, that Angles in same segment are equal.∠CBD =∠CAD ∠CAD =70° ∠BAD =∠BAC +∠CAD =30° + 70° = 100° ∠BCD+∠BAD=180° (Opposite angles of a cyclic quadrilateral) ∠BCD + 100° = 180° ∠BCD =180° - 100° ∠BCD=80° In ΔABC AB = BC (given) ∠BCA =∠CAB(Angles opposite to equal sides of a triangle) ∠BCA =30° also, ∠BCD = 80° ∠BCA +∠ACD =80° 30° + ∠ACD = 80°∠ACD =50° ∠ECD =50° Hence,∠BCD =80° & ∠ECD =50° thanks

eral whose diagonals intersect at a point E. If 4 DBCABCD is a cycliBAC is 30°, find ZBCD. Further, ifAB -BC, find 4 ECD.70°6.lic quadrilateral are diameters of the circle through the vertices of

Answer»

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

Angles in the same segment of a circle areequal.

For chord CD,

We know, that Angles in same segment are equal.∠CBD =∠CAD ∠CAD =70°

∠BAD =∠BAC +∠CAD =30° + 70° = 100°

∠BCD+∠BAD=180°

(Opposite angles of a cyclic quadrilateral)

∠BCD + 100° = 180°

∠BCD =180° - 100°

∠BCD=80°

In ΔABC

AB = BC (given)

∠BCA =∠CAB(Angles opposite to equal sides of a triangle)

∠BCA =30°

also, ∠BCD = 80°

∠BCA +∠ACD =80°

30° + ∠ACD = 80°∠ACD =50°

∠ECD =50°

Hence,∠BCD =80° & ∠ECD =50°

thanks