http://s6.photobucket.com/albums/y222/wing_hing/?action=view¤t=80ce57be.jpg
In the picture ,△XYZ is an equilateral triangle. M is the mid-point of YZ.
(a) Show that △XYM全等△XZM.
(b) Is XM a median, perpendicular bisector, altitude and angle bisector of △XYZ? Explain your answer.
中二 MATHS
2009-01-01 8:18 am
回答 (1)
2009-01-01 8:58 am
✔ 最佳答案
(a)XY = XZ (sides of equilateral Δ)
XM = XM (common)
YM = ZM (given that M is the mid-point of YZ)
ΔXYM ≡ ΔXZM (SSS)
(b)
M is the mid-point of YZ (given)
XM is a median.
ΔXYM ≡ ΔXZM (proved)
XMY = XMZ (corr. s of cong. Δs)
XMY + XMZ = 180o (adj. s on a st. line)
Thus, XMY = XMZ = 90o
XM ^ YZ
XM is an altitude.
XMY = XMZ = 90o (proved)
YM = ZM (given)
Thus, XM is a perpendicular bisector.
ΔXYM ≡ ΔXZM (proved)
YXM = ZXM (corr. s of cong. Δs)
Thus, XM is an angle bisector.
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