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 = 70^{o}

∠BAD = ∠BAC + ∠CAD

= 30^{o} + 70^{o}

= 100^{o}

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

∠BCD + 100^{o} = 180^{o}

∠BCD = 80^{o}

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°

**∠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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