Abcd is a rhombus in which altitude from d

  1. ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus.
  2. Solve the following :ABCD is a rhombus such that the perpendicular bisector of AB passes through D . Find the angles of the rhombus . Hint : Join BD . Then Δ ABD is equilateral.


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

Let sides of a rhombus be AB = BC = CD = DA = x Now, join DB. In ΔALD and ΔBLD, ∠DLA = ∠DLB = 90° AL = BL = `x/2`......[Since, DL is a perpendicular bisector of AB] And DL = DL......[Common side] ∴ΔALD ≅ΔBLD......[By SAS congruence rule] AD = BD......[By CPCT] Now, In ΔADB is an equilateral triangle. ∴∠A = ∠BDC = ∠ABD = 60° Similarly, ΔDBC is an equilateral triangle. ∴∠C = ∠BDC = ∠BC = 60° Also, ∠A = ∠C ∴∠D = ∠B = 180°– 60° = 120° .....[Since, sum of interior angles is 180°]

Solve the following :ABCD is a rhombus such that the perpendicular bisector of AB passes through D . Find the angles of the rhombus . Hint : Join BD . Then Δ ABD is equilateral.

Given that ABCD is a Rhombus and DE is the altitude on AB then AE = EB. In a ΔAED and ΔBED, DE = DE ( common line) ∠AED = ∠BED ( right angle) AE = EB ( DE is an altitude) ∴ ΔAED ≅ ΔBED ( SAS property) ∴ AD = BD ( by C.P.C.T) But AD = AB ( sides of rhombus are equal) ⇒ AD = AB = BD ∴ ABD is an equilateral traingle. ∴ ∠A = 60° ⇒ ∠A = ∠C = 60° (opposite angles of rhombus are equal) But Sum of adjacent angles of a rhombus is supplimentary. ∠ABC + ∠BCD = 180° ⇒ ∠ABC + 60°= 180° ⇒ ∠ABC = 180° - 60° = 120°. ∴ ∠ABC = ∠ADC = 120°. (opposite angles of rhombus are equal) ∴ Angles of rhombus are ∠A = 60° and ∠C = 60° , ∠B = ∠D = 120°.

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