Q. 63.9( 245 Votes )

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If DBC = 70°, BAC is 30°, find BCD. Further, if AB = BC, find ECD.

Answer :


METHOD 1:
For chord CD,

CBD = CAD (Angles on the same segment are equal)


CAD = 70o


BAD = BAC + CAD


= 30o + 70o


= 100o


BCD + BAD = 180° (Opposite angles of a cyclic quadrilateral)


BCD + 100o = 180o


BCD = 80o

Now, In ΔABC,


AB = BC (Given)


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


BCA = 30°


We have,


BCD = 80°


BCA + ACD = 80°


30° + ACD = 80°


ACD = 50°

As ACD = ∠ECD


∠ECD = 50°
METHOD 2:
On chord AD,
∠ABD = ∠ECD.....(1)  [Since angles on the same segment are equal]
In ΔABC,
Given: AB = BC
Therefore,
∠BAC = ∠BCA = 30º
Sum of angles of a triangle = 180º
30º + 30º + ∠ABC = 180º
∠ABC = 120º.....(2)
And we can see that
∠ABC = ∠ABD + ∠DBC
120º = ∠ABD + 70º
∠ABD = 50º
And from equation 1, we can see that,
∠ECD = 50º

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