Question

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

Answer 1

Given that ABCD is a Rhombus - Promotional Kites 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 triangle.∴ ∠A = 60°⇒ ∠A = ∠C = 60° (opposite angles of rhombus are equal)But Sum of adjacent angles of a rhombus is supplementary.∠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°.

Answer 2

Answer:

Step-by-step explanation:

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