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in a rectangle ABCD, E is a point  which bisects BC, PROVE that AE = ED.

Asked by nisha_vini29 13th September 2015, 12:41 AM
Answered by Expert
Answer:

C o n s i d e r space t h e space f o l l o w i n g space f i g u r e.

L e t space t h e space d i m e n s i o n s space o f space t h e space r e c tan g l e comma space b e space b r e a d t h equals a comma space a n d space l e n g t h equals b L e t space E space b e space t h e space m i d p o i n t space o f space B C space a n d space h e n c e comma space B E equals E C equals b over 2 N o w comma space t r i a n g l e s space A B E space a n d space D C E space a r e space r i g h t space t r i a n g l e s. T h u s comma space b y space a p p l y i n g space P y t h a g o r a s space T h e o r e m comma space w e space h a v e comma A B squared plus B E squared equals A E squared space a n d space C D squared plus E C squared equals D E squared space rightwards double arrow a squared plus open parentheses b over 2 close parentheses squared equals A E squared space a n d space a squared plus open parentheses b over 2 close parentheses squared equals D E squared space rightwards double arrow A E equals E D equals square root of a squared plus open parentheses b over 2 close parentheses squared end root rightwards double arrow A E equals E D equals square root of fraction numerator 4 a squared plus b squared over denominator 4 end fraction end root

Answered by Expert 14th September 2015, 6:01 PM
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