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Question
In Fig. ABCD is a quadrilateral in which AB = BC. E is the point of intersection of the right bisectors of AD and CD. Prove that BE bisects ∠ABC.
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Solution
Given: A quadrilateral ABCD in which AB = BC. PE and QE are right bisectors or AD and CD respectively such that they meet at E.
To prove: BE bisects ∠ABC.
Construction: Join AE, DE and CE.
Proof: Since, PE is the right bisector of AD and E lies on it.
∴ AE = ED ...(i)
[∵ Points on the right bisector of a line segment are equidistant from the ends of the segment]
Also, QE is the right bisector of CD and E lies on it.
∴ ED = EC ...(ii)
From (i) and (ii), we get
AE = EC ...(iii)
Now, in Δs ABE and CBE, we have
AB = BC ...[Given]
BE = BE ...[Common]
and AE = EC ...[From (iii)]
So, by SSS criterion of congruence
ΔABE = ΔACE
⇒ ∠ABE = ∠CBE
⇒ BE bisects ∠ABC.
Hence, BE is the bisector of ∠ABC.
Hence proved.
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