this is probably extreamly basic, but there a lot of questions so it might be a big burden. sorry about that
in a pyramid (SABCD) that its base is a rhombus (ABCD), all dihedral angles between the base and each face are 60 degrees, the (plumb line(?)) From S to ABCD lands on O.
prove that O is the center of the circle BLOCKED by ABCD, and therefore is in the intersection between AC and BD,
i've learned a lot about these types of questions in a very short time, so i got some information mixed up, would appreciate if someone cleared some things up for me:
1.1 how do you prove this? can it be proved by 1.2.1 OR 1.2.3 (which is my best shot) and even if so is there a better way?
1.2.1 in a circle that is blocked by a quadrilateral, there are 4 lines coming out of one point inside the quadrilateral that are perpendicular to each side of the perpendicular and all are equals, aka the blocked circle radiuses
if i find a point in which this happens inside a quadrilateral that CAN block a circle, that means that it is necessarily the blocked circle's center? do i have to prove this? this obviously happens here as ,FO=MO=LO=EO, just dont know how to prove O is in the intersection of BD and AC
1.2.2 if in a quadrilateral AC and BD are not equal, it means that SO cannot land in the intersection between them and if they are equal it means that means that the centers of the blocking and blocked circle are in the same place? (regardless of SO)
1.2.3 assuming BC||AD, AB||CD, since OM=OE and the angle between MO to BC is 90 and the same with OE to BC, is it OK to say that based on this information alone, O HAS TO BE ON the angle Angle bisector LINE between BC and CD? and on the Angle bisector LINE between BC and AD? and because its a rhombus then it has to be on the BD AND AC intersection? what happens when BD is not on the Angle bisector LINE?
the only doubt i have here is in the premise that if from a certain point Q there 2 EQUAL lines QM QN to 2 other lines AB AC and they both have 90 angle--(QM IS 90 degree into AB and same with QN AC) then Q has to be on the angle Angle bisector LINE between them
1.3 just to make sure, a pyramid in which its base is a triangle, if it can/cant be blocked by a circle/block a circle, does it means that SO will land in the center of the blocking/blocked circle? also same thing about a base which is a quadrilateral
2.1. lets say its the same exercise but with a trinangle, the angle between each face and the base is the same, how do you prove that SO lands in the center of the blocked circle? and does it imply somthing about the center of the blocking circle?
2.2 same thing but with blocking circle (ABC), in a blocking circle OA= OB=OC.
if i manage to prove that the landing point of S on ABC, lets call it Q applies QA=QB=QC, will it prove that Q is in the center of the blocking circle?
2.3 can you prove both 2.1 and 2.2 by working with the special traits that the center of blocked and blocking circle has with the triangle?
for instance a blocked circle has its center on the intersection between the 3 Angle bisector of the triangle, how can i prove it using that?
- last question, when they say that the angles between the face and the base are equal, does it have any implication on where (center of blocking, blocked circle) SO lands regardless of the shape of the base? also does it have an implication on the angle between (in the drawing) SD,SC.. and the base? in triangles or quadrilateral?
sorry for so much questions, just trying to get the hang of it
thanks in advance
Draw from $O$ the perpendicular lines to the sides of the base, and let their feet be $E$, $F$, $L$, $M$ as in your figure. From the Theorem of three perpendiculars it follows that $SE\perp CD$, $SF\perp AB$, $SL\perp AD$, $SM\perp BC$. But then, by hypothesis, $\angle SEO=\angle SFO=\angle SLO=\angle SMO=60°$, hence: $$ OE=OF=OL=OM=OS\cot60°. $$ This proves that $O$ is the center of the circle inscribed in the base.
Triangles $AOF$ and $AOL$ are equal by the hypotenuse-leg theorem, hence $\angle FAO=\angle LAO$. But we also have $\angle FAC=\angle LAC$ (from the equality of triangles $ABC$ and $ADC$). Hence point $O$ lies on line $AC$. In the same way one proves that $O$ lies on line $BD$.