197927001-การให้เหตุผล

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1 หน วยการเร ยนร       1 เร  อง การให เหตผลเก ยวก บร ปสามเหล   ยมและร ปส  เหล  ยม 1. ความร      นฐานเก   ยวก บการให เหต ผลทางเรขาคณต  ประโยคเง  อนไข  ในช ตประจ าว นเราม กจะพบข อความท  กษณะเป นประโยคเง  อนไข เช 1. ABCD เป นร ปส เหล   ยมม มฉาก แล วด านตรงข ามของ ABCD ความยาวเท าก  2. าเส นตรงสองเส นต ดก แล วม มตรงข ามม ขนาดเท าก  3. a เป นจ านวนค    แล a 2  เป นจ านวนค    ประโยคเง  อนไขด งกล าวม แบบเด ยวก คอ ประกอบด วยข อความสองข อความท  เช อมด วย ... แล ... เรยกข อความท   ตามหล  ว เหต  และเร ยกข อความท  ตามหล แล ผล อย างช ดเจน เช านวนน บท   หารด วย 2 ลงต เปนจ านวนค   สามารถน ามาเข ยนให อย   ในร ปประโยค .. แล ..  ได เป าจ านวนน บใดหารด วย 2 ลงต แล วจ านวนน บน   นเป นจ านวนค   บทกล บของประโยคเง   อนไข  จากประโยคเง  อนไข ปสามเหล   ยมสองร ปม ขนาดของม มเท าก นเป นค  ๆสามค   แล สามเหล ยมสองร ปเป นร ปสามเหล   ยมท  คล ายก าเราน าผลของประโยคเง  อนไขน   มาเป นเหต และน าเหต ของประโยคเง   อนไขน   มาเป นผล เราจะได บทกล บของประโยคเง อนไขเป นประโยคเง  อนไขใหม งน    ปสามเหล  ยมสองร ปเป นร ปสามเหล   ยมท  คล ายก แล ปสามเหล  ยมสองร ปน   นม ขนาดของ มเท าก นเป นค  ๆสามค   เราสามารถเข ยนประโยคเง  อนไขและบทกล บของประโยคเง อนไขข างต นให เป นประโยคเด ยวก  โดยใช คาว อเม   งน    ปสามเหล  ยมสองร ปม ขนาดของม มเท าก นเป นค  ๆสามค   อเม    ปสามเหล  ยมน   นม านยาว เท าก นสองด านเม  อม ประโยคท  เช  อมด วย อเม  เราสามารถเขยนประโยคน   นเป นประโยคเง อนไขสองประโยค เช ปสามเหล  ยมใดเป นร ปสามเหล ยมหน าจ  กต อเม  ปสามเหล   ยมน   นม านยาวเท าก นสองด าน สามารถเข ยนได เป  าร ปสามเหล   ยมใดเปนรปสามเหล ยมหน าจ   แล วรปสามเหล  ยมน   นมด านยาวเท าก นสองด าน และ าร ปสามเหล   ยมใดม านยาวเท าก นสองด าน แล วร ปสามเหล   ยมน   นเป นร ปสามเหล   ยมหน าจ   

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    1. ABCD ABCD 2. 3. a a2

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    1.1

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    1) 2) ABCD ABCD 3) 4) 5)

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    2)

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    4) 180

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    2

    1. 2.

    1 ABCD ABC

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    A B

    C D

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    180

    180

    2 360

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    A B

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    1.2

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    1 = 4 3 + 2 = 180

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    3 4

    EF AB CD E F AE X = DF Y AB CD

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    AB // CD GH AB CD E F 1) GE A = DF H 2) GE B + CF E = 180

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    AB // CD DE BC E BE D = AB E + ED C

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    2.

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    2.1

    1. AC BD M AB // DC

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    2. ABC DBC BC AB = AC DB = DC AD ABD ACD

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    3. ABC AB = AC CD BE AB AC D E CD = BE AD = AE

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    4. ABC A = B = C ABC

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    5.

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    2.2

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    .......................................................................................................................................... 2. AB CD O ABCD

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    3. ABCD E F AD BC EB DF DFBE

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    4.

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