Question

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 1

Answer:

Solution:

Consider the chord CD,

As we know, angles in the same segment are equal.

So, ∠CBD = ∠CAD

∴ ∠CAD = 70°

Now, ∠BAD will be equal to the sum of angles BAC and CAD.

So, ∠BAD = ∠BAC + ∠CAD

= 30° + 70°

∴ ∠BAD = 100°

As we know, the opposite angles of a cyclic quadrilateral sums up to 180 degrees.

So,

∠BCD + ∠BAD = 180°

Since, ∠BAD = 100°

So, ∠BCD = 80°

Now consider the ΔABC.

Here, it is given that AB = BC

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

∠BCA = 30°

also, ∠BCD = 80°

∠BCA + ∠ACD = 80°

So, ∠ACD = 50° and,

∠ECD = 50°

Hope it will be helpful :)

Answer 2

Answer:

arcs is called a segment the circle.

Angles in the same segment of a circle are equal.

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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°

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