Geometry proof problems for mathematics urgency speed good addition

It is easy to prove that ead = 90 degrees, (to prove that ead is half of a flat angle) and that there are three quadrilaterals with 90 degrees of angles, of course, are rectangular.

Prove that EAD is half of the square angle: the perpendicular line at the base of the isosceles triangle is also the angular bisector of the bac, and an is also the angular bisector of the outer corner.

When ABC is a right triangle, a quadrilateral ADCE is a square.

Because it is an isosceles right triangle so dca=45 degrees, because adc=90 degrees, so dac=45 degrees, and the two angles are equal, so it is an isosceles triangle ad=dc

Rectangles with equal margins are squares, which is proven.

cam= abc+abc, because ab=ac, so abc=acb, because cn is the abc outer angle of the bisector of cam, so mae= cae, so cae= acb, so ae cd, so ead= dce=90°, quadrilateral adce is rectangular.

When ABC is an isosceles right triangle, the quadrilateral ADCE is a square. Because ABC is an isosceles right triangle, AD=CD, so the quadrilateral ADCE is rectangular.

Square the two sides of the inequality and it can be proved.

a +b )+b +c )+a +c ) pat the town 2)*(a+b+c)2(a +b +c)+2[( A Sui a +b )+b +c )+a +c attack rough)] 2(a +b +c )

The above formula is true, so the original formula is true.

According to the sum of the internal angles of the parallelogram is equal to 360 degrees and the theorem that the parallelogram is equal to the diagonals

a+∠aef=180°；d+ deg=180° is known a=120°, so.

aef=180°-120°

According to the theorem of equality of internal wrong angles between parallel lines, it is introduced:

efg=∠aef

It is known that d=100°, so.

deg=180°-100°

According to the theorem of equality of internal wrong angles between parallel lines, it is introduced:

egf=∠deg

It is introduced according to the theorem that the sum of the internal angles of a triangle is equal to 180°.

gef+∠egf+∠efg=180°

So gef=180°- egf- efg=180°-60°-80°

Solution: Because ab ef, aef=60Because cd eg, d=100°. So aeg=100

So, gef= aeg- aef=40. So ged=80. Because of AD BC

So efg= aef=60° , egf= ged=80

The first step is to prove that the two planes bcc1b1 and add1a1 are parallel (you can make the perpendicular bf perpendicular to ad, and then prove that ad is perpendicular to the two planes of two intersecting straight lines, so that the parallel is obtained from the parallel of two planes perpendicular to the same line), and the second step is to obtain ec and ad parallel according to the intersection of the surface acd intersecting the two parallel planes (look at that theorem).

The second question is that you can change the vertex and treat C-A1AB as A1-ABC. That is, A1 is the vertex, ABC is the bottom surface, and the base area is directly multiplied by the height.

After a rough look, the first question proves that bcb1c1 is parallel AA1D1D, then the line is in the plane, the line is parallel to the surface, and finally the line is parallel.

The second question is my method, extend ab, dc to complete the graph, turn it into a triangular prism, and then conduct the height of the same volume.

Time is tight, I don't know if I can make it, LZ try.

Solution: (1) If the radius is equal, the angle of the arc is equal, the arc is equal, and the angle AD+BC=90=BC, then the corresponding arc is also equal.

2) No, the OE angle doc is two angles, where doe=aod, then eoc=boc, connect de, ec, then ad+cb=de+ec, in the triangle dec, de+ec>cd, so it is not true.

Because in the isosceles trapezoidal ABCD, AOB = 60 , the triangle doc and triangle AOB are both regular triangles.

Connect CR, because R is the midpoint of DO, so the angle CRB=90 degrees, i.e. the triangle CRB is a right triangle, and Rq is the hypotenuse midline, so RQ=1 2BC

In the same way, by connecting bp, we get pq=1 2bc

And the P point is the AO midpoint, so rp=1 2AD, and AD=BC so rp=rq=pq

from'Whereas, the triangle PQR is an equilateral triangle.

Prove that the three sides are equal.

Connect CR, BP

AOB and DOC are both equilateral triangles.

Then bp is vertical ac, cr is vertical do.

The triangle CRB, BPC is a right triangle, and the middle line of the hypotenuse of the right triangle is equal to half of the hypotenuse.

ar=1/2bc=ap

Then prove rp=1 2ad=1 2bc

Then the three sides are equal and proven.