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標題:
中二 MATHS
發問:
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.
最佳解答:
(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. =
中二 MATHS
發問:
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.
最佳解答:
(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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