Plane Geometry
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A circle passes through 6 points A, F, H, B, C and G, as shown in the diagram. D and B are 2 points on AE such that AD = DB = BE = DF, GCE and FDC are straight lines
Prove that triangle ADF is congruent to triangle CDB (no prob with this part)
Prove that triangle GEA is similar to triangle CEB (need help)
Surprisingly, I can prove that triangle GEA is similar to triangle BEC but not that required in the question. If possible, will someone help me out with this? Thanks!
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gad. u got the 3 angles just see the angles match or not la, if not = qn ask wrong or u prove wrong la =.=
sorrying for insulting u .
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In triangle GEA and triangle CEB
angle GEA = Angle CEB (same angle)
angle AGE= Angle CBE (external angle of a cyclic quadrilateral)
Therefore triangle GEA is similar to triangle CEB (AA property)
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Originally posted by SBS n SMRT:
In triangle GEA and triangle CEB
angle GEA = Angle CEB (same angle)
angle AGE= Angle BCE (external angle of a cyclic quadrilateral)
Therefore triangle GEA is similar to triangle CEB (AA property)
sorry arh, but i think angle age = bce only if ag and bc are parallel, which you have no evidence of. but then i also no solution, sorry to TS
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Originally posted by yiha093:
gad. u got the 3 angles just see the angles match or not la, if not = qn ask wrong or u prove wrong la =.=
sorrying for insulting u .
your answer is like telling ppl either you are right or you are wrong, seems not very useful
dun get angry with me okay?
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Originally posted by MasterMoogle:
sorry arh, but i think angle age = bce only if ag and bc are parallel, which you have no evidence of. but then i also no solution, sorry to TS
SBS is correct.
Assume AGE = x degree
Therefore ABC = 180 - x (angle in opp segment/Opp angle of cyclic quad are supp)
CBE = 180 - ABC ( angle on str line)
= 180-(180-x)
= x
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Originally posted by Mikethm:
SBS is correct.
Assume AGE = x degree
Therefore ABC = 180 - x (angle in opp segment/Opp angle of cyclic quad are supp)
BCE = 180 - ABC ( angle on str line)
= 180-(180-x)
= x
how do u get BCE = 180 - ABC?
they not on the same line wor, and ae obviously not // to ge
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Originally posted by MasterMoogle:
how do u get BCE = 180 - ABC?
they not on the same line wor, and ae obviously not // to ge
ABE is a straight line. " B and D are points on the line AE" the question explicitly said.
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Originally posted by MasterMoogle:
your answer is like telling ppl either you are right or you are wrong, seems not very useful
dun get angry with me okay?
its okay, just showing TS my logic.
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Originally posted by Mikethm:
ABE is a straight line. " B and D are points on the line AE" the question explicitly said.
aiyaya, i dun mean that.
i mean why u minus angle bce and abc when they are on different lines
tks anyway
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Originally posted by MasterMoogle:
aiyaya, i dun mean that.
i mean why u minus angle bce and abc when they are on different lines
tks anyway
Opposite angles of cyclic quad are supplementary.ABCG is a cyclic quad. I made a typo.
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i made some adjustment to my answer. sorry.
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Hi!
While I agree that all the proving done here is correct, may i ask if there's a need to follow the "order" of the triangles angles given? (eg. in the question they ask to provethat triangle GEA is similar to triangle CEB)
If that's the case, you said angle AGE = angle CBE but if you rearrange the above question, angle AGE should be equal to angle BCE.
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Originally posted by yiha093:
gad. u got the 3 angles just see the angles match or not la, if not = qn ask wrong or u prove wrong la =.=
sorrying for insulting u .
LOL. Nice solution, indeed. -
Originally posted by anpanman:
Hi!
While I agree that all the proving done here is correct, may i ask if there's a need to follow the "order" of the triangles angles given? (eg. in the question they ask to provethat triangle GEA is similar to triangle CEB)
If that's the case, you said angle AGE = angle CBE but if you rearrange the above question, angle AGE should be equal to angle BCE.
the question seems to have a error.
should be AGE is similar to CBE and not what the qns says,