The ABC triangle inscribed the circle center . I is the mid point of BC . M is a random point on IC ( M differ from C and D ) . D is the intersection of the circle center O and AM . the intersection of the tangent to the circle of a triangle AMI in M with BD , BC in P and Q.
1 ) prove that : DM . IA = MP .IB
2 ) calculating : MP/MQ = ?
The ABC triangle inscribed the circle center . I is the mid point of BC . M is a random point on IC ( M differ from C and D ) ...
#1
Posted 24-04-2016 - 10:28
- leminhnghiatt and hoakute like this
#2
Posted 26-04-2016 - 11:19
hinhf 3.png
The ABC triangle inscribed the circle center . I is the mid point of BC . M is a random point on IC ( M differ from C and D ) . D is the intersection of the circle center O and AM . the intersection of the tangent to the circle of a triangle AMI in M with BD , BC in P and Q.
1 ) prove that : DM . IA = MP .IB
2 ) calculating : MP/MQ = ?
1)
We have $\widehat{AMP} =\widehat{AMI} +\widehat{IMP}$
$=\widehat{AMI} +\widehat{IAM} =\widehat{AIB}$(1)
$\Leftrightarrow\widehat{PMD} =\widehat{AIC}$ (2)
have $\widehat{MDP} =\widehat{ICA}$ (3)
from (2, 3) then $\triangle PMD\sim\triangle AIC$(angle-angle)
$\Rightarrow\frac{MD}{IC} =\frac{MP}{IA}$
$\Rightarrow MD .IA =MP .IC =MP .IB$ (4)
2)
(1)$\Rightarrow\widehat{DMQ} =\widehat{BIA}$ (5)
have $\widehat{MDQ} =\widehat{IBA}$ (6)
from (5, 6) then $\triangle MDQ\sim\triangle IBA$ (angle-angle)
$\Rightarrow\frac{MD}{IB} =\frac{MQ}{IA}$
$\Rightarrow MD .IA =MQ .IB$ (7)
from(4, 7) then $MP .IB =MQ .IB$
$\Leftrightarrow MP =MQ$
- PlanBbyFESN and Sangnguyen3 like this
(Giúp với Tính $\int_m^n\left(\sqrt{ax^4 + bx^3 + cx^2 + dx + e}\right) dx$)
(Tam giác ABC cân tại A, lấy D trên cạnh BC, r1,r2 là bán kính nội tiếp ABD, ACD. Xác định vị trí D để tích r1.r2 lớn nhất )
(Nhấn nút "Thích" thay cho lời cám ơn, nút Thích nằm cuối mỗi bài viết, đăng nhập để nhìn thấy nút Thích)
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