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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