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In the figure above, D is a point on side AC of ?ABC.

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In the figure above, D is a point on side AC of ?ABC. [#permalink]

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11 Apr 2005, 17:10
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18. In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosceles?
(1) The area of triangular region ABD is equal to the area of triangular region DBC.
(2) BD┴AC and AD = DC
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11 Apr 2005, 19:43
b too

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12 Apr 2005, 05:12
b is the OA... care to explain? I thought D
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12 Apr 2005, 18:14
B...
Trian ABD and Trian CBD are congruent triangles by the Side- angle-side theorem and hence AB is equal to BC

Angle ADC = Angle CDB --- Angle
Bd is the common side --- Side

Hope this helps

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13 Apr 2005, 03:30
ritledge wrote:
18. In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosceles?
(1) The area of triangular region ABD is equal to the area of triangular region DBC.
(2) BD┴AC and AD = DC

"B". but how come i can't see any figure above or my eyes are playing tricks on me
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13 Apr 2005, 07:32
I guess he has forgotten to post the figure ..... even i was

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03 Jul 2006, 04:45
swath20 wrote:
I guess he has forgotten to post the figure ..... even i was

I think both B suggests that the triangle can be issoceles or equilateral.

B is a theorem in an issoceles triangle whereas the same theorem is valid for equilateral triangle too.

The answer should be E ? Need some discussion on this one.

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03 Jul 2006, 04:45
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