[Toán 7] Bài toán Chứng minh về Tam giác

C

chuotbachkute

H

hiensau99

scaled.php


a, + Gọi Ax là tia đối của tia AC

+ Ta có $\widehat{BAC}+\widehat{BAx}= 180^o$ (kề bù) $\to \widehat{BAx} = 180^o -\widehat{BAC}$

+ AE là tia phân giác $\widehat{BAx} \to \widehat{A_1}=\widehat{A_2}= \dfrac{180^o -\widehat{BAC}}{2}= 90^o - \dfrac{\widehat{BAC}}{2}$

+ $\Delta ABC$ có $\widehat{BAC}+ \widehat{ABC}+ \widehat{C} = 180^o \to \widehat{BAC} = 180^o - \widehat{ABC} - \widehat{C} $

+ $ \widehat{ABC}$ là góc ngoài ở đỉnh B của $\Delta ABC$ nên:

$ \widehat{ABC}= \widehat{A_1}+ \widehat{E}= 90^o - \dfrac{\widehat{BAC}}{2}+ \widehat{E} = 90^o - \dfrac{180^o - \widehat{ABC} - \widehat{C}}{2}+ \widehat{E} $

$= 90^o - 90^o + \dfrac{ \widehat{ABC}}{2}+ \dfrac{ \widehat{C}}{2}+ \widehat{E}= \dfrac{ \widehat{ABC}}{2}+ \dfrac{\widehat{C}}{2}+ \widehat{E}$

$\to \widehat{E} = \widehat{ABC} + \dfrac{ \widehat{ABC}}{2}+ \dfrac{ \widehat{C}}{2} = \widehat{ABC} - \dfrac{ \widehat{ABC}}{2}- \dfrac{ \widehat{C}}{2} = \dfrac{ \widehat{ABC}}{2}- \dfrac{ \widehat{C}}{2}$

Vậy $\widehat{E} = \dfrac{ \widehat{ABC}}{2}- \dfrac{ \widehat{C}}{2}$ (đpcm)

b, Ta có : $BK // AE \to \widehat{A_2}= \hat{K_1} $ (đồng vị) và $ \widehat{A_1}= \hat{B_1}$ (so le trong)

Mà $ \widehat{A_2}= \hat{A_1} \to \hat{B_1} = \hat{K_1}$ (đpcm)


jfkrebvfđkmvbksjrhgvmd cvmdfnhgk
dmfhnjmhgkjm xmbfkxdhgjnb v
dsmghkjnbvc,mghtuom
 
You must log in or register to reply here.
Top Bottom