Question

Consider four collinear points A, B, C and D in that order. Another point E is such that AB = BE = BC. If DE2 = AD. CD, then find the measure (in degrees) of (∠CED + ∠BEA + ∠AEC – ∠CBE).

Answer 1

Given :  Consider four collinear points A, B, C and D in that order. Another point E is such that AB = BE = BC.

DE² = AD.CD

To Find :  ∠CED + ∠BEA + ∠AEC – ∠CBE

Solution:

AB = BE = BC

=> ΔABE & ΔCBE are isosceles right angle triangle

=> ∠BAE = ∠AEB = ∠CEB = ∠BCE = 45°

  ∠AEC = ∠CBE = 90°

DE² =  BD² + BE²

DE² = AD.CD  = (BD + AB). (BD - BC) =  (BD + BE). (BD - BE)

= BD² - BE²

BD² + BE² =  BD² - BE²

=> BE = 0   Hence AB = BC = 0  

Looks mistake in Data

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