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Question
ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus.
Solution
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,
Then, ΔADB is an equilateral triangle.
∴ ∠A = ∠ADB = ∠ABD = 60°
Similarly, ΔDBC is an equilateral triangle.
∴ ∠C = ∠BDC = ∠DBC = 60°
Also, ∠A = ∠C
∴ ∠D = ∠B = 180° – 60° = 120° ...[Since, sum of interior angles is 180°]
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