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ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus.

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Answer: 120^{\circ},120^{\circ},60^{\circ},60^{\circ}

Solution.
Let sides of a rhombus be AB = BC = CD = DA = x
Join DB

Here  \angle DLA=\angle DLB=90^{\circ}             
  {DL is perpendicular bisector of AB}

AL=BL=\frac{X}{2}
DL = DL {common side of \triangle ADL and \triangle BDL}
 \triangle ALD\cong \triangle BLD              {by SAS congruence}
AD = BD

In \triangle ADB,  AD = AB = DB = x
 \triangle ADB is an equilateral triangle.
\therefore \angle ADB=\angle ABD=\angle A=60^{\circ}
Similarly, \triangle DCB is an equilateral triangle
\angle BDC=\angle DBC=\angle C=60^{\circ}
Also,
\angle A = \angle C
 \angle D = \angle B          …..(1)

\angle A+ \angle B+\angle C+\angle D=360^{\circ}

60^{\circ}+ B+60^{\circ}+B=360^{\circ}
2\angle B=\frac{240}{2}=120^{\circ}
Hence answer is 120^{\circ},120^{\circ},60^{\circ},60^{\circ}

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