In △ ABC, D and E are on the lines AB and AC respectively, ad = 2, BD = 5, EA = 4, EC = 10, Ed: BC=
AD/AB=2/(2+5)=2/7
AE/AC=4/(4+10)=4/14=2/7
Ad / AB = AE / AC, so de ‖ BC
AD /AB=DE/BC=2/7
Known: as shown in the figure, in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is ()
A. 30°B. 36°C. 45°D. 50°
Let ∠ EBD = x °, be = De, EDB = ∠ EBD = x °, AED = ∠ EBD + ∠ EDB = 2x °, ad = De, a = ∠ AED = 2x °, BDC = ∠ a + ∠ abd = 3x °, BD = BC, C = ∠ BDC = 3x °, AB = AC, ABC = ∠ C = 3x °, a + ∠ ABC + ∠ C = 180 °, 2x +
RT triangle ABC, ∠ ACB = 90 °, ab = 5cm, AC = 4cm, CD ⊥ AB in D?
In △ ACB and △ CDB, ∠ ABC = ∠ CBD. ∠ CDB = ∠ ACB = 90 ° because the sum of internal angles of triangles is equal, ∠ a = ∠ BCD. So △ ACB is similar to △ CDB. So CB: ab = CD: AC, CB ^ 2 = AB ^ 2-ac ^ 2, that is CB = 3. So 3:5 = CD: 4, the solution is: CD = 12 / 5 (12 / 5)
In RT triangle ABC, the angle c = 90 ° if a: B = 3:4, C = 10, then a =
A=6 B=8
It is known that, as shown in the figure, △ ABC, point E is on the central line BD, ∠ DAE = ∠ abd
It is proved that: (1) ∵ - DAE = - abd, ∵ - ADE = - BDA, ∵ △ ade ∵ BDA. (2 points) ∵ - addd = dead, (2 points) & nbsp; that is, ad2 = de · dB. (1 point) (2) ∵ D is the midpoint on the AC side, ∵ ad = DC. ∵ addd = dead, ∵ dcbd = dedc, (2 points) ∵ - CDE = - BDC. (1 point)