As shown in the figure, in triangle ABC, angle c is equal to 90 degree ad bisector angle BAC, if BC = 64 and BD: CD = 9; 7 Then the distance between D and ab is (to process) (what is the distance between D and ab

As shown in the figure, in triangle ABC, angle c is equal to 90 degree ad bisector angle BAC, if BC = 64 and BD: CD = 9; 7 Then the distance between D and ab is (to process) (what is the distance between D and ab

Passing point D makes de ⊥ AB in E
∵BD；CD＝9：7,BC＝64
∴CD＝7/（9+7）×BC＝7/16×64＝28
∵ ad bisection ∠ BAC, ∠ C = 90, de ⊥ ab
■ de = CD = 28 (property of angular bisector)
The distance between point D and ab is 28
(the distance from point d to AB is the length of the line segment crossed by D and perpendicular to ab)

If O is any point in the triangle ABC, the intersection of Bo and AC is extended to e. it is proved that ab + AC > ob + OC

It is proved that Bo is extended and AC is handed over to point E
From "the difference between the two sides of a triangle is less than the third side", it can be concluded that
BE-AB＜AE
OC-OE＜CE
∵BE=OB+OE
∴OB+OE-AB＜AE
OC-OE＜CE
Add the above two expressions to get the
OB-AB+OC＜AE+CE
That is ab + AC > ob + OC

Point O is a point outside the triangle ABC. Take a ′, B ′, C ′ on the ray OA, ob, OC respectively, and connect a ′, B ′, B ′, C ′, a ′,
Let a 'B' / / AB, B 'C' / / BC, a 'C' / / AC. is the triangle a 'B' C 'similar to ABC? Please prove your conclusion

According to the similarity theorem
(3) If the three sides of a triangle are proportional to the three sides of another triangle, then the two triangles are similar
Because a'b '/ / AB, B'c' / / BC, a'c '/ / AC
So OA: OA '= AB: a' B '= AC: a' C '
OB:OB′=AB:A′B′=BC:B′C′
OC:OC′=CB:C′B′=AC:A′C′
That is ab: a ′ B ′ = AC: a ′ C ′ = BC: B ′ C ′