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HUNG

Cho tam giác ABC vuông tại A có góc B=55 độ.Trên nửa mặt phẳng không chứa B , kẻ tia Cx vuông góc với AC.Trên tia Cx lấy điểm D sao cho CD=AB.

a. Tính số đo góc ACB

b. Chứng minh 

ΔABC=ΔCDA ;AD song song với BC.

c. Kẻ AH vuông góc với BC (H thuộc BC) và CK vuông góc với AD(K thuộc AD).Chứng minh BH=CK

d. Gọi I là trung điểm của AC.Chứng minh ba điểm I,K,L thẳng hàng

Giúp mik thực hiện bài toán này nhé!!!!!

Edogawa Conan
9 tháng 7 2019 lúc 10:33

A B C D I H K

a) Xét t/giác ABC có \(\widehat{A}\) = 900

=> \(\widehat{B}+\widehat{C}=90^0\) 

=> \(\widehat{C}=90^0-\widehat{B}=90^0-55^0=35^0\)

b) Xét t/giác ABC và t/giác CAD

có : AB = CD (gt)

  \(\widehat{BAC}=\widehat{ACD}=90^0\) (gt)

   AC : chung

=> t/giác ABC = t/giác CAD (c.g.c)

=> \(\widehat{BCA}=\widehat{CAD}\) (2 góc t/ứng)

Mà 2 góc này ở vị trí so le trong

=> AD // BC

c) Xét t/giác HAB và t/giác KCD

có: \(\widehat{BHA}=\widehat{CKD}=90^0\) (gt)

     AB = CD (gt)

   \(\widehat{B}=\widehat{D}\) (vì t/giác ABC = t/giác CDA)

=> t/giác HAB = t/giác KCD (ch - gn)

=> BH = KD (2 cạnh t/ứng) (xem lại đề)

d)  Ta có: BH + HC = BC

 AK + KD = AD

Mà BH = KD (cmt); BC =AD (vì t/giác ABC = t/giác CDA)

=> HC = AK

Xét t/giác AIK và t/giác CIH

có: AI = IC (gt)

  \(\widehat{KAI}=\widehat{ICH}\)(vì t/giác ABC = t/giác CDA)

  AK = CH (cmt)

=> t/giác AIK = t/giác CIH (c.g.c)

=> \(\widehat{AIK}=\widehat{HIC}\)(2 góc t/ứng)

Mà \(\widehat{AIH}+\widehat{HIC}=180^0\)(kề bù)

hay \(\widehat{AIK}+\widehat{AIH}=180^0\)

=> ba điểm H, I, K thẳng hàng (xem lại đề)

HUNG
10 tháng 7 2019 lúc 8:07

Cảm ơn bạn Edogawa Conan , mình được 9 điểm nhé ! :)

Trần Hiếu Anh
9 tháng 3 2020 lúc 14:59

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