# The angle between the altitudes of a parallelogram,

Question:

The angle between the altitudes of a parallelogram, through the same vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.

Solution:

Draw a parallelogram $\mathrm{ABCD}$.

Drop a perpendicular from $\mathrm{B}$ to the side $\mathrm{AD}$, at the point $\mathrm{E} .$

Drop perpendicular from $\mathrm{B}$ to the side $\mathrm{CD}$, at the point $\mathrm{F}$.

In the quadrilateral $\mathrm{BEDF}:$

$\angle \mathrm{EBF}=60^{\circ}, \angle \mathrm{BED}=90^{\circ}$

$\angle \mathrm{BFD}=90^{\circ}$

$\angle \mathrm{EDF}=360^{\circ}-\left(60^{\circ}+90^{\circ}+90^{\circ}\right)=120^{\circ}$

In a parallelogram, opposite angles are congruent and adjacent angles are supplementary.

In the parallelogram $\mathrm{ABCD}$ :

$\angle \mathrm{B}=\angle \mathrm{D}=120^{\circ}$

$\angle \mathrm{A}=\angle \mathrm{C}=180^{\circ}-120^{\circ}=60^{\circ}$

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