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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.

Mathematics
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Other answers:

1...show that triangle ABD=triangle ACD......it gives BE=EC 1...show that triangle ABE=triangle AEC......it gives
I have shown that triangle ABD = ACD, but I don't know how to do the other one.
triangle ABE=triangle AEC because 1...AC=AB 2...BE=EC 3...AE=AE
But we have to prove that BE = EC right?
Perpendicular bisector.
u've shown that triangle ABD=triangle ACD am i right?
yes.
sorry if I take a little long to reply.
np....:)
well it gives BE = EC
But how?
I mean, its not a side of ABD or ACD.
u r right ... thats not a complete proof... lets think again...
we need to show that triangle ABE=triangle AEC...........
try to find 2 angles equal...and one side equal for that 2 triangle
AB = AC
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so that you don't need to scroll up too much.
lol.........:)
we proved that triangle ABD=triangle ACD so that...
AE is common so triangle AEB = AEC.
we are done
But how does that prove that AD is the perpendicular bisector?
We should prove that AD bisects BC and angle AEB = 90 degrees.
it provides BE=EC and
Thanks a lot mukushla!!!
yw......:)

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