pratu043
ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.
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pratu043
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|dw:1344485886935:dw|
mukushla
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|dw:1344486523841:dw|
pratu043
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ok
mukushla
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1...show that triangle ABD=triangle ACD......it gives BE=EC
1...show that triangle ABE=triangle AEC......it gives <AEB=<AEC=90
pratu043
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I have shown that triangle ABD = ACD, but I don't know how to do the other one.
mukushla
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triangle ABE=triangle AEC because
1...AC=AB
2...BE=EC
3...AE=AE
pratu043
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But we have to prove that BE = EC right?
pratu043
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Perpendicular bisector.
mukushla
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u've shown that triangle ABD=triangle ACD
am i right?
pratu043
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yes.
pratu043
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sorry if I take a little long to reply.
mukushla
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np....:)
mukushla
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well it gives BE = EC
pratu043
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But how?
pratu043
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I mean, its not a side of ABD or ACD.
mukushla
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u r right ... thats not a complete proof... lets think again...
mukushla
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we need to show that triangle ABE=triangle AEC...........
mukushla
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try to find 2 angles equal...and one side equal for that 2 triangle
pratu043
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AB = AC
pratu043
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|dw:1344488387156:dw|
pratu043
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so that you don't need to scroll up too much.
mukushla
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lol.........:)
mukushla
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we proved that triangle ABD=triangle ACD so that... <BAE=<CAE
and what is the third condition?
pratu043
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AE is common so triangle AEB = AEC.
mukushla
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we are done
pratu043
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But how does that prove that AD is the perpendicular bisector?
pratu043
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We should prove that AD bisects BC and angle AEB = 90 degrees.
mukushla
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it provides BE=EC and
<AEB=<AEC but we know that <AEB+<AEC=180
pratu043
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Thanks a lot mukushla!!!
mukushla
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yw......:)