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Geometry Problem URGENTTTTTT

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Geometry Problem URGENTTTTTT [#permalink] New post 10 Nov 2011, 07:54
Q1 .In this diagram, AB=AC. and BD=CD. Which of the following statements is true?
A. BE=EC
B. AD is perpendicular to BC
C. triangle BDE and CDE are congruent
D. angle ABD equal angle ACD
E. All of these

Figure: SEE ATTACHMENT

Please give explanation for each of these why something is not true

Q2. ABCD is a rectangle; the diagonals AC and BD intersect at E. Which of the following statements is not necessarily true?

A. AE=BE
B. Angle AEB equals angle CED
C. AE is perpendicular to BD
D. Triangles AED and AEB are equal in area
E. Angle BAC equals angle BDC

Pls give full explanation
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Re: Geometry Problem URGENTTTTTT [#permalink] New post 11 Nov 2011, 16:02
E / C

Q1.

AB=AC, BD=CD
-> ABC, DBC are isosceles triangles respectively.
-> Angle ABD = Angle ACD
-> Triangle ABD = Triangle ACD
-> Angle BAD = Angle CAD
-> BC is perpendicular to AD (because Triangle ABC is isosceles.)

-> Triangel ABE=ACE, DBE=BCE,
and Angle AEB=AEC=DEB=DEC=90
A. BE=EC <- true, Triangel AEB=AEC
B. AD is perpendicular to BC <- true
C. Triangel BDE=CDE <- true
D. Angle ABD=ACD <- true
E. all true <- true

answer : E


Q2.

Rect.ABCD -> AB=CD, AD=BC,
and, according to the symmetricalness of rect., AE=BE=CE=DE
-> tri. AEB=CED : isosceles,
tri. BEC=AED : isosceles.

A. AE=BE <- true
B. ang. AEB=CED <- true; vertical angles.
C. AE perpend. BD <- not necessarilly; true only when the rect. is a square.
D. tri. AED and AEB are equal in area <- true; these 2 triangles are 2 parts of tri. ABD. When you assume BE and DE are the bases of each tri., they are sharing the height. So, they have the same height and the same length of the base, and thus the same area.
E. ang. BAC=BDC <- true; tri. EAB=EDC.

answer : C
Re: Geometry Problem URGENTTTTTT   [#permalink] 11 Nov 2011, 16:02
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