geometryquadrilateral
$ABCD$ is cyclic quadrilateral.The side $AB$ is extended to $E$ in such a way that $BE =BC$.If $\angle ADC =70$ degree,$\angle BAD=95$ degree then $\angle DCE$ is … ?
Note:$ABCD$ is cyclic the sum of two opposite angles equals $\pi$.
$\angle BCD=85$ degree, $\angle ABC=110$ degree so $\angle DCE=180-x-85$ degree.
Observe Triangles $ACB$ and $ADC$.
$\angle ACB$ is common. And $AB=CD$, $AC=AC$(Common).
Therefore the triangles are similar (Two sides have lengths in the same ratio, and the angles included between these sides have the same measure).
Which means,
$\frac{AB}{AC}=\frac{AC}{AB}$, and the angle between them is common( $\angle ACB$)
Now,
$\angle CAD = \angle ACB =c$
Thus, $a=c$
Therefore,
$5a=180^0$. $a=36^0$ and $\angle BAC=72^0$.
I have four different and one general solution to this problem, I hope it will help you.
Best Answer
As ABCD is a cyclic quadrilateral and opposite angles sum up to π: ∠BAD=95 => ∠BCD=85 ∠ADC=70 => ∠ABC=110
Also as in triangle BEC, ∠CBE=70 (Since, ∠ABC + ∠CBE = π (a straight line)); and as BE=BC; the opposite angles to both these sides must be equal; i.e, ∠BCE=∠BEC; & ∠BCE + ∠BEC = π - 70; => ∠BCE=∠BEC=55
Thus, ∠DCE = ∠BCD + ∠BCE = 85 + 55 = 140