Ex.10.5 Q6 Circles Solution - NCERT Maths Class 9

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\(\begin {align} {ABCD}\end {align}\) is a cyclic quadrilateral whose diagonals intersect at a point \(\begin {align} {E. }\end {align}\)If \(\begin {align} \angle {DBC}=70^{\circ}, \quad \angle {BAC}=30^{\circ} \end {align}\) find \(\begin {align} \angle {BCD}. \end {align}\) Further if \(\begin {align} {AB=BC,} \end {align}\) find \(\begin {align} \angle {ECD.} \end {align}\)

 Video Solution
Ex 10.5 | Question 6

Text Solution

What is known?

\(\begin {align} {ABCD}\end {align}\) is cyclic quadrilateral. \(\begin {align} {DBC}=70^\circ \end {align}\) and \(\begin {align} {BAC}=30^\circ. \end {align}\) \(\begin {align}{AB = BC.}\end {align}\)

What is unknown?

Value of \(\begin {align} \angle {BCD} \text { and } \angle {ECD} \end {align}\)


  • A quadrilateral \(\begin {align} {ABCD}\end {align}\) is called cyclic if all the four vertices of it lie on a circle.
  • The sum of either pair of opposite angles of a cyclic quadrilateral is \(180^\circ.\)
  • Sum of angles in a triangle is \(180^\circ.\)
  • Angles in the same segment are equal.


Based on the data given, draw the figure.

In the triangles \({ABD} \) and \({BCD,} \) \(\begin {align} \angle {CAD}=\angle {CBD}=70^{\circ}. \end {align}\) (Angles in the same segment)

So \(\begin {align} \angle {BAD}=30^{\circ}+70^{\circ}=100^{\circ} \end {align}\)

Since \(\begin {align} {ABCD }\end {align}\)is a cyclic quadrilateral, the sum of either pair of opposite angles of cyclic quadrilateral is \(180^{\circ}. \)

\[\begin{align} \angle {BAD}+ \angle {BCD}&=180^{\circ} \\ \angle {BCD} &=180^{\circ}-100^{\circ} \\&=80^{\circ} \end{align}\]

Also given \(\begin{align} {AB = BC.}\end{align}\)

So, \(\begin {align} \angle {BCA}=\angle {BAC}=30^{\circ} \end {align}\) (Base angles of isosceles triangle are equal)

\[\begin{align} ∴ \quad \angle {ECD} &=\angle {BCD}-\angle {BCA} \\ &=80^{\circ}-30^{\circ} \\ &=50^{\circ} \end{align}\]

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