It is known that A1B1C1 ABC is a regular triangular prism and D is the midpoint of AC. Verify AB1 pl

Anonymous |2012-10-25

As shown in the figure, it is known that in the triangular prism ABC-A1B1C1, D is the AC midpoint. Verify: ab1 plane dbc1 no picture, but can not be drawn I describe it is to draw a triangular prism and then the top is a1b1c1 and the bottom is abc to connect ab1 fast!

I'll answer. Recommended answers.

To complete the parallelepiped abec ab1e1c1, take the midpoint of b1e1 as d1.

Abec Ab1E1C1 is a parallelepiped, Ab C1E1, AC E1D1, Bad C1E1D1.

d and d1 are ac and d1 respectively

Connect B1C, cross B1 to E, connect De

The quadrilateral b1bcc1 is rectangular

then b1e=ec

In ab1c, AD=DC, DE AB1, while DE Planar DBC1ABC1 DBC1

I also just made it, maybe my method is not the best.

Make ab midpoint e, BB1 midpoint f, make BC1 midpoint g, connect ed, fe, fg, dg

e and f are the midpoints, respectively.

ef∥ab1

D and E are the midpoints of ac and ab, f and g are the midpoints of bb1 and bc1, respectively, and b1c1 = bc

fg = 1 2b1c1 = 1 2bc = ed, and fg 1 2b1c1 1 2bc ed

fgde is a parallelogram.

EF dgab1 ef,dg on planar dbc1.

ab1 planar dbc1

A1B1C1-ABC is a regular triangular prism, and the quadrilateral B1Bcc1 is a rectangle connecting B1C, and crossing B1 to E, then B1E=EC connects De

In ab1c, ad=dc, de ab1, and ab1 6 5-plane dbc1 de 6 3-plane dbc1

ab1 dbc1 I'm Yang Hao.

Take the midpoint of a1c1 as d1 and connect ad1 and b1d1

Because AD is parallel and equal to D1C1, AD1 is next to DC1, and it is easy to know that D1B1 is parallel to DB, so the surface AD1B1 is parallel to the surface DB1

After the most radical index, since the line ab1 is on the surface AD1B1, the ab1 plane DBC1 is certified.

Connect B1C, let B1C cross B1 to point F, and connect DF

Because abc—a1b1c1 is a regular triangular prism, b1bcc1 is rectangular, so the point f is the midpoint of b1c.

So df is a median line of the triangle ACB1, so df is parallel to ab1, and df is in plane dbc1, so ab1 is parallel to plane db1

Take the midpoint of a1c1 as d1 and connect ad1 and b1d1

Because AD is parallel and equal to D1C1, AD1 is parallel to DC1, and it is easy to know that D1B1 is parallel to DB, so the face AD1B1 is parallel to the surface DBC1.

Finally, since the line ab1 is on the surface AD1B1, the ab1 plane DBC1 is proven.

or through C1 as a parallel line of B1D1 to E1, C1E1 is equal to 3 2 times B1D1; take the midpoint e of db; B1 is parallel to B1F and is equal to A1C1; Connect EE1, FC; It is easy to know that CF is parallel and equal to ab1, the face DBE1C1 is coplanar with the surface DBC1, since EC is parallel and equal to E1F, so EE1 is parallel and equal to CF (i.e., AB1), because EE1 is on the surface DBE1C1, so AB1 plane DBC1 is proved.

Connect B1C, let B1C cross B1 to point F, connect Df Because Bc—A1B1C1 is a regular triangular prism, so B1Bcc1 is rectangular, so Point F is the midpoint of B1C.

So df is a median line of the triangle ACB1, so df is parallel to ab1, and df is in plane dbc1, so ab1 is parallel to plane db1

In a regular triangular prism, the midpoint F of BC is connected to AF, then AF is perpendicular to the plane BB1C1C, so AF is perpendicular to FC1

So point f is point d.

Connected to DE, AA1 is parallel and equivalent to DE, so AD is parallel to A1EA1E planar ADC1