Question

Q. 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

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Answer 2

Given :-

• ${\sf ABCD \: is \: a \: cyclic \: quadrilateral}$

• ${\sf It's \: diagonals \: intersect \: at \: Point \: E}$

• ${\sf \angle DBC = 70 \degree}$

• ${\sf \angle BAC = 30 \degree}$

To Find :-

• ${\sf \angle BCD }$

• ${\sf If \: AB = BC \: . Evaluate \: \angle ECD}$

Solution :-

Before starting the solution , we shall know that ;

• The Sum of Opposite angles of a cyclic quadrilateral is $\bf 180 \degree$

• The angles in same segment are always equal

• Angle opposite to equal sides are equal or vice versa.

__________________________

Now , see the attachment 1 ;

Here we have ;

• ${\sf \angle DBC = 70 \degree\quad \{ \because Given\}}$

• ${\sf \angle BAC = 30 \degree\quad \{ \because Given\}}$

Now here ;

${: \implies \quad \sf \angle DBC = \angle DAC \quad \{ \because In \: same \: segment \}}$

${: \implies \quad \sf \angle DAC = 70 \degree \{ \because \angle DBC = 70 \degree \}}$

Now , Also ;

${: \implies \quad \sf \angle BCD + \angle BAD = 180 \degree }$

${: \implies \quad \sf \angle BCD + ( \angle BAC + \angle DAC) = 180 \degree }$

${: \implies \quad \sf \angle BCD + ( 30 \degree + 70 \degree ) = 180 \degree }$

${: \implies \quad \sf \angle BCD + 100 \degree = 180 \degree }$

${: \implies \quad \sf \angle BCD = 180 \degree - 100 \degree }$

${: \implies \quad \bf \therefore \quad \angle BCD = 80 \degree }$

Now see the Attachment 2 ;

In ${\sf \triangle ABC }$

• AB = BC

Which implies that ;

${: \implies \quad \sf \angle BAC = \angle BCA \quad \{ \because \: Angles \: opposite \: to \: AB \: and \: BC\}}$

${: \implies \quad \sf \angle BCA = 30 \degree \quad \{ \because \: \angle BAC = 30 \degree \}}$

Now , Also ;

${\quad \leadsto \quad \sf \angle ECD + \angle BCA = \angle BCD}$

Put the values ;

${: \implies \quad \sf \angle ECD + 30 \degree = 80 \degree }$

${: \implies \quad \sf \angle ECD = 80 \degree - 30 \degree }$

${: \implies \quad \bf \therefore \quad \angle ECD = 50 \degree }$

Now , we are Done :D