Triangular pyramid P ABC mid angle ABC 90 degrees, PA PB PC proves that the planar PAC is perpendicu

Evidence: Take the midpoint M of AC and connect PM, because PA=PC, PAC is an isosceles triangle, and PM is the midline of PAC, so PM is perpendicular to AC. If BM is connected, there is AM=BM, because PA=PB, PM=PM, so PAM is all equal to PBM, so PMA= PMB=90°, that is, PM is vertical BM.

Therefore, PM is perpendicular plane ABC, and PM belongs to the plane PAC, so the plane PAC is perpendicular to the plane ABC.

If P is in the plane of the triangle abc, where should it be? It should be the midpoint of the AC side, so that Pa=Pb=PC and P is a point outside the plane of the triangle ABC, then P can only move on the perpendicular bisector of the AC perpendicular to the surface ABC. There are several verifications of this idea.

Evidence: Take the midpoint M of AC and connect PM, because PA=PC, macro hunger pac is an isosceles triangle, and PM is the midline of PAC, so PM is perpendicular to ACConnect BM, then there is AM=BM, because Yuxi is PA=PB, PM=PM, so PAM is all equal to PBM, so PMA= PMB=90°, that is, PM vertical BM

Therefore, the PM vertical flat town is absolutely the surface ABC, and the PM belongs to the plane PAC, so the plane PAC is the vertical plane ABC

∠acb=90°

ac⊥bc ①

Plane PBC is perpendicular to plane ABC and plane PBC Plane ABC=BC is obtained by the infiltration hall AC plane PBC (two planes are perpendicular, then the straight line perpendicular to the intersection line in one plane must be perpendicular to the other plane hall).

and PB planar PBC

PB vertical shouting is purely AC

After A as AD PB to D

The surface of the PAB surface PBC, and the PB is the intersection of the surface PAB and the surface PBC, and the AD PB, the AD surface PBC, obtains: AD BC

PA Face ABC, Mishu PA BC

Acre is dressed as ad bc, pa bc, and pa ad a, bc face pab, bc ab

PA vertical bottom surface ABC launches PA perpendicular to BC---1 plane PAB vertical plane PBC, then A does AD perpendicular to PBad perpendicular to PBa, perpendicular to slow hand surface PBC

then AD is perpendicular to BC---2

By 1 2 get the royal is BC vertical disturbance demolition suspicion straight to the face of the PAB so AB vertical BC

After A as AD PB to D

The surface of the surface is the surface PBC, and the PB is the intersection of the surface PAB and the surface PBC, and the AD PB, and the AD surface PBC, obtain: AD BC.

PA surface ABC, PA BC.

AD BC, PA BC, while PA AD A, BC Facial PAB, BC AB.

Solution: pa surface abc, pa=ab=ac=2 pa ab, pa ac

Then PAB and PAC are isosceles right triangles.

pb=pc=2√2

Pass E as EF PB to F and connect AF

Then ef is parallel and equal to 1 2pb = 2

ae is the midline of the hypotenuse of the isosceles right triangle PAC = 1 2pc = 2af is the midline of the bottom edge of the isosceles triangle with a vertex angle of 60°.

af⊥bc, ∠fac=60°/2=30°∴af=accos30°=√3

cos∠aef=(ae²+ef²-af²)/(2*ae*ef)=(2+2-3)/(2*2)=1/4

The cosine of the angle formed by the heteroplane straight line ae and pb is: 1 4

In RT pac, E is used as EG AC to G PA plane ABC, and eg plane ABC

eg=1/2pa=1

s△abc=1/2ab*absin∠bac=1/2*2*2*√3/2=√3

v a-ebc=1/3* s△abc*eg=√3/3

Go to the midpoint of ab, m be, the midpoint n, the midpoint of pb, and connect to mn mq qn

man for the desired horn.

qm= mn= an= 7 2 cosine theorem shows cos = 2 7 7

It's a simple question! You can use geometry to solve it, and the problem is very simple. Create a Cartesian coordinate system!

Take the midpoint O of BC, connect AO, PO, BAC=90°, AB=AC=1, ABC is isosceles RT, BC= 2, AO= 2 2, according to the inverse theorem of the Pythagorean theorem, BPC=90°,PO BC=2 2, the same reason, PAO is also isosceles RT, PO AO, AO=O, PO Plane ABC, AO is the projection of PA in plane ABC, PA=45°, the angle between PA and the bottom ABC is 45 degrees.

The perpendicular line of the bottom ABC is Po, O is the perpendicular foot, because Pa=PB=PC, so OA=ob=OC

and bac is the right angle, so o is the midpoint of the hypotenuse bc. Even AO, then Pao is the angle between PA and the bottom ABC.

It is easy to find ao=bc 2= 2 2

So cos pao=ao pa= 2 2, pao 45°

It's 60°.

Proof: Since Pa=PB=PC, the projection of the point P on the bottom ABC should be the outer center of the triangle ABC, because A=90°, then the point H is the midpoint of BC, that is, the angle Pah is the angle formed by the straight line Pa and the bottom ABC, considering that the triangle PAH is a right-angled triangle, and AH=(1 2)AB=(1 2)Ah, then the angle PAH=60°

Take the midpoint F of the cavity AC, connect EF and BF, let PA=AB=AC=1, then EF is the median line of PAC, EF PA, EF=PA 2=1 2, FEB is the angle formed by PA and BE, PA Flat Bridge Deck ABC, EF Plane ABC, BF Plane ABC, EF BF, in RT ABF, AF=1 2, AB=1, according to the Pythagorean theorem, AB 2+AF 2=BF 2, BF= 5 2, the same is true according to the Pythagorean theorem, BE= 6 2, the cosine formed by the angle of cospa and be is 6 6. Leak this fiercely.