Question
ABCD is a parallelogram. E is a point on CD such that BE bisects ∠ABC. BE = 12 cm, CE = 19 cm, ∠BEC = ∠AED. DE = ___ cm.
Answer:
:
∠ABE = ∠BEC (alternate angles)
∠AED = ∠EAB
∠AED = ∠BEC
∠EAB = ∠EBA
As BE bisects ∠ABC
∠EBC = ∠ABE
From (1) and (2) as the two angles of the triangle ABE are equal to two angles of triangle BEC, the two triangles are similar
So, BECE=ABBE
BE2= (CE + ED)CE
Substituting BE = 12cm and CE = 9cm, we get ED = 7cm
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:
∠ABE = ∠BEC (alternate angles)
∠AED = ∠EAB
∠AED = ∠BEC
∠EAB = ∠EBA
As BE bisects ∠ABC
∠EBC = ∠ABE
From (1) and (2) as the two angles of the triangle ABE are equal to two angles of triangle BEC, the two triangles are similar
So, BECE=ABBE
BE2= (CE + ED)CE
Substituting BE = 12cm and CE = 9cm, we get ED = 7cm
Was this answer helpful ?
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