a: Xét ΔCBA vuông tại A và ΔCDA vuông tại A có

AB=AD

AC chung

DO đó: ΔCBA=ΔCDA

Suy ra: \(\widehat{ACB}=\widehat{ACD}\)

hay CA là tia phan giác của góc BCD

b: Xét ΔCHA vuông tại H và ΔCKA vuông tại K có 

CA chung

\(\widehat{HCA}=\widehat{KCA}\)

Do đó: ΔCHA=ΔCKA

Suy ra: CH=CK

c: Xét ΔCDB có 

CH/CD=CK/CB

DO đó; HK//DB

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)

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

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Trả lời:

-O : góc chung

-OA = OC

-OB = OD

=> tam giác OAD = tam giác OCB

b/ Xét tam giác ACD và tam giác CAB có

-AC: cạnh chung

-OA = OC

OB = OD

⇒⇒AB = CD

-AD = CB (vì ΔΔOAD=ΔΔOCB)

Vậy tam giác ACD = tam giác CAB

                                              ~Học tốt!~

a) Xét \(\Delta DAO\) và \(\Delta BCO\)có:

OA=OC(gt)

\(\widehat{O}\) chung

OB=OD (dt)

=> \(\Delta DAO=\Delta BCO\left(g.c.g\right)\)

=> CB=AD (2 cạnh tương ứng)

Xét \(\Delta ABC\) và \(\Delta CDA\)có:

CB=AD (cmt)

AB=CD (OA=OC, OB=OD)

AC chung

=> \(\Delta ABC=\Delta CDA\left(ccc\right)\)

a: Xét ΔABC có \(BC^2=AB^2+AC^2\)

nên ΔABC vuông tại A

b: Xét ΔBAE vuông tại A và ΔBDE vuông tại D có

BE chung

\(\widehat{ABE}=\widehat{DBE}\)

Do đó: ΔBAE=ΔBDE

Suy ra: BA=BD; EA=ED

c: Xét ΔAEK vuông tại A và ΔDEC vuông tại D có

EA=ED

\(\widehat{AEK}=\widehat{DEC}\)

Do đó:ΔAEK=ΔDEC

Suy ra: EK=EC

lollolol là j

\(\frac{\sqrt[2]{4}\cdot8\text{go 06}}{5^2+7tx\%}=\infty\cdot7?\)

2√4·8go 0652+7tx% =∞·7?