Câu 10:
Given the isosceles triangle ABC (AB=AC) with A =108 . Draw the bisector AD and BE of angles A and B respectively. Given BE = 10cm. Evaluate AD.
Answer: AD = ? cm.
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Given the isosceles triangle ABC (AB=AC) with \(A=108^o\). Draw the bisector AD and BE of angles A and B respectively. Given BE = 10cm. Evaluate AD.
Given the isosceles triangle ABC (AB=AC) with . Draw the bisector AD and BE of angles A and B respectively. Given BE = 10cm. Evaluate AD.
Given the isosceles triangle ABC (AB=AC) with . Draw the bisector AD and BE of angles A and B respectively. Given BE = 10cm. Evaluate AD.
câu hỏi thế này thì bố thằng nào hiểu được ?????
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29 tháng 3 2018 lúc 16:41
Given an isosceles triangle ABC (AB = AC), \(\widehat{A}\) = 108o. AD and BE are the bisectors of angle A and B, BE = 10cm. Caculate AB.
In triangle ABC, BC=AC and BCA=900. D and E are points on AC and AB respectively such that AD=AE and 2CD =BE.Let P be the point of intersection of BD with the bisector of angle CAB. What is the angle PCB in degrees?
In triangle ABC, BC=AC and BCA=900. D and E are points on AC and AB respectively such that AD=AE and 2CD =BE.Let P be the point of intersection of BD with the bisector of angle CAB. What is the angle PCB in degrees?
1. Two bisector BD and CE of the triangle ABC intersect at O. Suppose that BD.CE 2BO.OC . Denote by H the point in BC such that .OH⊥BC . Prove that AB.AC 2HB.HC 2. Given a trapezoid ABCD with the based edges BC3cm , DA6cm ( AD//BC ). Then the length of the line EF ( Ein AB,Fin CD and EF // AD ) through the intersection point M of AC and BD is ............... ? 3. Let ABC be an equilateral triangle and a point M inside the triangle such that MA^2MB^2+MC^2 . Draw an equilateral triangle ACD whe...
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1. Two bisector BD and CE of the triangle ABC intersect at O. Suppose that BD.CE = 2BO.OC . Denote by H the point in BC such that .\(OH⊥BC\) . Prove that AB.AC = 2HB.HC
2. Given a trapezoid ABCD with the based edges BC=3cm , DA=6cm ( AD//BC ). Then the length of the line EF ( \(E\in AB,F\in CD\) and EF // AD ) through the intersection point M of AC and BD is ............... ?
3. Let ABC be an equilateral triangle and a point M inside the triangle such that \(MA^2=MB^2+MC^2\) . Draw an equilateral triangle ACD where \(D\ne B\) . Let the point N inside \(\Delta ACD\) such that AMN is an equilateral triangle. Determine \(\widehat{BMC}\) ?
4. Given an isosceles triangle ABC at A. Draw ray Cx being perpendicular to CA, BE perpendicular to Cx \(\left(E\in Cx\right)\) . Let M be the midpoint of BE, and D be the intersection point of AM and Cx. Prove that \(BD⊥BC\)
Given the right triangle ABC (A^ = 90o), BD is the bisector of the angle at B ( D of AC ). If AD = 6cm and AB = 12cm then the area of the right triangle ABC is ...... cm2.
I don't know English very much so i can't answere your question. Sory about that :(
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Given the right triangle ABC (A^ = 90o), BD is the bisector of the angle at B ( D of AC ). If AD = 6cm and AB = 12cm then the area of the right triangle ABC is ...... cm2.
Mới học lớp 7 thôi, ko làm được bài nhưng để mk dịch đề thử nhá:
Cho tam giác ABC (A^= 90o), BD là tia phân giác của góc B (D thuộc AC). Nếu AD= 6cm, AB= 12cm thì diện tích của tam giác ABC là .....cm2.
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Dựng DE vuông góc với BC (E thuộc BC).2 tam giác vuông ABD và EBD bằng nhau (gcg) nên \(S_{ABD}=S_{EBD}=\frac{1}{2}\cdot12\cdot6=36\)(cm2)Và cạnh DE=AD=6(cm)2 tam giác vuông CED và CAB có chung góc C thì đồng dạng với nhau với tỷ lệ cạnh là: \(\frac{DE}{AB}=\frac{6}{12}=\frac{1}{2}\)nên tỷ lệ diện tích bằng bình phương tỷ lệ cạnh, ta có:\(\frac{S_{CED}}{S_{CAB}}=\frac{1}{2^2}=\frac{1}{4}\)Vậy \(S_{ABC}=S_{ABD}+S_{BDE}+S_{CED}=\left(2+\frac{1}{4}\right)S_{ABD}=\frac{9}{4}S_{ABD}=\frac{9}{4}\cdot36=81\)(cm2).
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