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# In the given figure, $ABCD$ is a Rhombus. If $\angle A = {80^ \circ }$, then find $\angle CDB$.

Last updated date: 15th Jun 2024
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Hint: First use the property of transversal line for interior angles to get the relation between the angle A and D. Then use the property of Rhombus that its diagonals bisect the angles to get the desired result.

We have given in the problem that $ABCD$ is a Rhombus and the measure of the given angle is $\angle A = {80^ \circ }$.
So, we can say that $AB||CD$ and all the sides are equal in length using the property of Rhombus.
As we have given that $AB||CD$, then the line $AD$ is the transversal line through these lines.
Using the property of transversal line, that the pair of interior angles on the same side of the transversal line is supplementary. So, we have
$\angle A + \angle D = {180^ \circ }$
$\Rightarrow$$\angle D = {180^ \circ } - \angle A Substitute the value of the angle \angle A = 80^\circ into the equation. \Rightarrow$$\angle D = {180^ \circ } - {80^ \circ }$
$\Rightarrow$$\angle D = {100^ \circ } We know that, \angle D can be express as: \Rightarrow$$\angle ADB + \angle BDC = \angle D$
As we know, that the diagonals of the Rhombus bisect the angles, so we can conclude that $BD$ dissect the angle $\angle D$, so we have$\angle ADB = \angle BDC$. Then,
$\Rightarrow$$2\angle BDC = {100^ \circ } \Rightarrow$$\angle BDC = \dfrac{{{{100}^ \circ }}}{2}$
$\Rightarrow$$\angle BDC = {50^ \circ } So, the required measure of the angle,\angle BDC = {50^ \circ }. Note: Alternative method: We know that \angle A = \angle C and \angle B = \angle D (opposite angle of the Rhombus are equal) We know that the sum of all the interior angles of the Rhombus is 360^\circ . So, we have \angle A + \angle B + \angle C + \angle D = 360 Substitute \angle A = \angle C and \angle B = \angle D into the equation: \Rightarrow$$2\angle A + 2\angle D = 360$
$\Rightarrow \angle A + \angle D = 180$
$\Rightarrow$$\angle D = {180^ \circ } - \angle A Now, we substitute the given angle,\angle A = {80^ \circ }into the equation: \Rightarrow$$\angle D = {180^ \circ } - {80^ \circ }$
$\Rightarrow$$\angle D = {100^ \circ } As we know, that the diagonals of the Rhombus bisect the angles, so we can conclude that BD dissect the angle \angle D, so we have \angle ADB = \angle BDC. Then, \Rightarrow$$2\angle BDC = {100^ \circ }$
$\Rightarrow$$\angle BDC = \dfrac{{{{100}^ \circ }}}{2} \Rightarrow$$\angle BDC = {50^ \circ }$
So, the required measure of the angle,$\angle BDC = {50^ \circ }$.