In triangle ABC, the bisector of AB = AC = 3 BC = 2 ∠ ABC intersects the parallel line of BC at d to find the area of triangle abd

In triangle ABC, the bisector of AB = AC = 3 BC = 2 ∠ ABC intersects the parallel line of BC at d to find the area of triangle abd


3,3,2 triangles determined
High square 3 * 3-1 * 1 = 8
Area: 2 * root 8 divided by 2
It's two root two



As shown in the figure, the bisector of the inner angle ∠ ABC of △ ABC intersects with the bisector of the outer angle ∠ ACG at point D, and the parallel line of BC crosses AB to e and AC to F through point D. try to judge the relationship between EF and be, CF, and explain the reason


EF = be-cf. prove: ∵ BD bisection ∵ ABC, ∵ abd = ∵ DBC. Also ∵ ed ∥ BC, ∵ EDB = ∵ DBC; ∵ abd = ∵ EDB, ∵ be = ed; similarly prove: CF = FD; ∵ EF = ed-fd, ∵ EF = be-cf



In △ ABC, the bisector of ∠ C intersects AB at D, and the bisector of BC intersects AC at e through D. given BC = a, AC = B, the length of De can be obtained


∵ de ∥ BC, ∥ 1 = ∥ 3. And ∥ 1 = ∥ 2, ∥ 2 = ∥ 3DE = EC, from △ ade ∥ ABC, ∥ DEBC = aeac, DEA = B − DEB, B ∥ de = AB-A · De, so de = ABA + B