In the given figure ,ABCD is a square and EF is parallel to diagnol DB and EM=FM.Prove that:(i)BF=DE,(ii)AM bisects angle BAD

Asked by abinash.gupta003 | 26th Aug, 2016, 04:13: PM

Expert Answer:

begin mathsize 14px style left parenthesis straight i right parenthesis
In space triangle BCD comma space BC equals CD
rightwards double arrow angle CBD equals angle CDB
Also comma space EF space is space parallel space to space BD comma
angle CEF equals angle CDB space and space angle CFE equals angle CBD space space space space left parenthesis corresponding space angles right parenthesis
rightwards double arrow angle CEF equals angle CFE
rightwards double arrow CE equals CF
rightwards double arrow CD minus DE equals BC minus BF
rightwards double arrow DE equals BF

left parenthesis ii right parenthesis
In space triangle ADE space and space triangle ABF comma
AD equals AB
angle ADE equals angle ABF space space space space space left parenthesis each space 90 degree right parenthesis
DF equals BE space space space space space left parenthesis From space left parenthesis straight i right parenthesis right parenthesis
So space triangles space ADE space and space ABF space are space similar space by space SAS space property
rightwards double arrow AE equals AF space
and space FM equals EM space left parenthesis given right parenthesis
AM equals AM
So space triangles space triangle AMF space and space triangle AME space are space similar space by space SSS space property
rightwards double arrow angle FAM equals angle EAM
rightwards double arrow angle FAM plus angle BAF equals angle EAM plus angle EAD
rightwards double arrow angle DAM equals angle BAM
rightwards double arrow AM space bisects space BAD end style

Answered by Rashmi Khot | 26th Aug, 2016, 05:08: PM

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