In the triangle ABC, the angle ACB is 90 degrees, and D is the intersection of the bisectors of the three interior angles of the triangle ABC. If AC = 3, BC = 4, find the distance from D to ab?

The bisector of the inner angle of a triangle is set at one point, and the distance to the three sides is equal. Therefore, the distance from D to ab can be set as X. according to the area series equation: 3x + 4x + 5x = 3 × 4, the solution is x = 1

As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 4cm, BC = 3cm, now fold △ ABC to make vertex A and B coincide, then the length of crease De is () cm
A. 52B. 154C. 158D. 5

Let AE = x, be = x, CE = 4-x, in RT △ BCE, ∵ BC2 + CE2 = be2, ∵ 32 + (4-x) 2 = X2, ∵ x = 258, in RT △ ade, ad = 52, AE = 258, so de = AE2 − ad2 = (258) 2 − (52) 2 = 158 (CM)

In RT △ ABC, if an acute angle of 30 ° is known, the other acute angle is ()

180°-90°-30°=60°

As shown in figure k-33-2, in the triangle ABC, D is the midpoint on the AB side, e is a point on the AC side, DF is parallel to be, EF is parallel to AB, and DF and EF intersect at point F
Verification: AE and DF are equally divided

DF is parallel to be, EF is parallel to AB, that is, EF is parallel to BD, so befd is a parallelogram. BD = EF, D is the midpoint of AB, ad = BD, so ad = EF. According to ef parallel to ad, EF = ad, we can know that two triangles are the same, and AE and DF can be bisected by the intersection

In the triangle ABC, the angle ACB is 90 degrees, and D is the intersection of the bisectors of the three interior angles of the triangle ABC. If AC = 3, BC = 4, find the distance from D to ab?