# Elementary geometry questions and answers

Recent questions in Elementary geometry
Brunton39 2022-06-23

### I'm curious about is there any geometric property relative to negative value for determinant of matrix.$det\left(A\right)<0$I knew about some of determinant of matrix properties as following, but it seems to me that it is nothing relative to negative value of determinant of matrix$det\left(AB\right)=det\left(A\right)det\left(B\right)\left(Multiplicative\right)$$det\left(A\right)=\mathit{\text{0}}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\mathit{\text{A is singular}}$${M}_{2,2}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$$|det\left({M}_{2×2}\right)|=|ad-bd|=\mathit{\text{volumn of parallelogram}}$

Armeninilu 2022-06-23

### Let H be the orthocenter of acute $\mathrm{△}ABC.$ . Points D and M are defined as the projection of A onto segment BC and the midpoint of segment BC, respectively. Let ${H}^{\prime }$ be the reflection of orthocenter H over the midpoint of DM, and construct a circle $\mathrm{\Gamma }$ centered at ${H}^{\prime }$ passing through B and C. Given that $\mathrm{\Gamma }$ intersects lines AB and AC at points $X\ne B$ and $Y\ne C$ respectively, show that points X, D, Y lie on a line.

telegrafyx 2022-06-22

### In $ABC$, $\mathrm{\angle }B<\mathrm{\angle }C$. If $D$ and $E$ are in $AC$ and $AB$ respectively, such that $BD$ and $CE$ are angle bisectors, prove that $BD>CE$.Well, $AC, but is there a formula for angle bisector length?

Cory Patrick 2022-06-22

### Finding explicit formula for intersection of perpendicular bisectors of a triangle

Devin Anderson 2022-06-22

### In the interior of a triangle ABC, a point P is marked in such a way that: $PC=BC$ and the measure of the angle PAB is equal to the measure of the angle PAC which is $17°$ . calculate the measure of angle PCB, if the measure of angle $B={107}^{o}$

Kyla Ayers 2022-06-22

### Let a circle $\omega$ (not labelled in the graph) centered at P tangent to AB, and T is point of tangency. $\mathrm{\angle }APB={90}^{\circ }$ . Let K (not labelled in the graph) be some point on the circle $\omega$ , the semicircle with diameter BK intersects PB at Q. Let R be the radius of that semi-circle. If $4{R}^{2}-AT\cdot TB=10$ and $PQ=\sqrt{2}$ , calculate BQ.

Jaqueline Kirby 2022-06-22

### How can I write the equation of a sphere that is centered at the triple point $\left(2,4,-4\right)$ and passes through the origin.

Nagambika 2022-06-21

### Which among the following is/are not a regular polygon

Finley Mckinney 2022-06-21

### The points (0,9),(12,0) and (0,0) create a right triangle. Find the equation of the angle bisectors for each of the three angles on the triangle.I am unsure of how to find the equations. Any help will be greatly appreciated!

Devin Anderson 2022-06-21

### From origin walk halfway to (8, 6), turn 90 degrees left, then walk twice as far.

Jaqueline Kirby 2022-06-21

### Here are the five postulates:1. Each pair of points can be joined by one and only one straight line segment.2. Any straight line segment can be indefinitely extended in either direction.3. There is exactly one circle of any given radius with any given center.4. All right angles are congruent to one another.5. If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if extended indefinitely, meet on that side on which the angles are less than two right angles.Questions:1. These to me sounds more like something that shouldn't require proving... does it?2. Why is it important to stress things that are obvious? For example, what other answers can you get when extending a line segment other than it can be extended indefinitely in either direction?3. Similarly, what space can allow two circle of the same radius and center to be not the same?4. Saying all right angles are congruent ... isn't that the same as saying all 64.506 degree angles are congruent? Isn't it ANY angle are congruent if they are the same degrees measures from the same reference point (say x-axis)?5. Why do we need the 5th postulate?

Sarai Davenport 2022-06-20

### Prove that angle bisectors of a triangle are concurrent using vectors. Also, find the position vector of the point of concurrency in terms of position vectors of the vertices.I solved this without using vectors to get some idea. I am not sure how to prove it using vectors. I don't want to use vector equations for straight lines and then find the point of concurrency. That's like solving using coordinate geometry.Let $\stackrel{\to }{a},\stackrel{\to }{b},\stackrel{\to }{c}$ represent the sides $A,B,C$ respectively.The angle bisectors are along $\frac{\stackrel{\to }{a}}{|\stackrel{\to }{a}|}+\frac{\stackrel{\to }{b}}{|\stackrel{\to }{b}|},\phantom{\rule{1em}{0ex}}\frac{\stackrel{\to }{b}}{|\stackrel{\to }{b}|}+\frac{\stackrel{\to }{c}}{|\stackrel{\to }{c}|},\phantom{\rule{1em}{0ex}}\frac{\stackrel{\to }{c}}{|\stackrel{\to }{c}|}+\frac{\stackrel{\to }{a}}{|\stackrel{\to }{a}|}$Let the sides $AB,BC,CA$ be $x,y,z$. Let $AD$ be one of the angular bisector.$\frac{BD}{CD}=\frac{x}{z}$Hence$D=\frac{x\stackrel{\to }{c}+z\stackrel{\to }{b}}{x+z}$What should be the next step? Or is there a better method?

Quintin Stafford 2022-06-20

### For reference: In the drawing, T is the point of tangency, $LN||AT$, $OH=4$ and $L{N}^{2}+A{M}^{2}=164$ . Calculate HN.

April Bush 2022-06-18

### This might be an easy question, but since i'm new to solid shapes, i couldn't solve it. I need to find m and n so that segments BD and AC have the smae midpoint.

Theresa Archer 2022-06-18

### I'm trying to get an overarching understanding of the components of mathematical systems so that in my self study of each category of math I can break them down by their unique aspects, i.e. the operators they use, the major concepts they deal with (i.e. how calculus is about "change"), etc.As far as my experience with formal math terminology goes, im rather weak, and i get utterly confused by the technicality required in formal definitions.As a good starting point, I'd like to better understand what the difference is between an axiom, a theorem and a postulate. At my current level of knowledge i would use them interchangeably (lol), however I'm sure one is founded upon the others.If someone could explain the logical hierarchy/relation between these three it would be greatly appreciated.

arridsd9 2022-06-17

### Is it true that:-an inner product satisfies the properties of a norm if and only if the norm satisfies the parallelogram equality-a norm can be induced by a metric if and only if the metric satisfies $d\left(x+a,y+a\right)=d\left(x,y\right)$ and $d\left(ax,ay\right)=ad\left(x,y\right)$or are the implications one way?

Kendrick Hampton 2022-06-17

### I've looked all over and I can't find a good proof of why the diagonals of a rhombus should intersect at right angles. I can intuitively see its true, just by drawing rhombuses, but I'm trying to prove that the slopes of the diagonals are negative reciprocals and its not working out.I'm defining my rhombus as follows: $\left[\left(0,0\right),\left(a,0\right),\left(b,c\right),\left(a+b,c\right)\right]$I've managed to figure out that $c=\sqrt{{a}^{2}-{b}^{2}}$ and that the slopes of the diagonals are $\frac{\sqrt{{a}^{2}-{b}^{2}}}{a+b}$ and $\frac{-\sqrt{{a}^{2}-{b}^{2}}}{a-b}$What I can't figure out is how they can be negative reciprocals of one another.I mean to say that I could not find the algebraic proof. I've seen and understand the geometric proof, but I needed help translating it into coordinate form.

minwaardekn 2022-06-16

### I'm given that the distance between two points in the Beltrami-Klein model is$d\left(XY\right)=\frac{1}{2}ln\left(\frac{\overline{XQ}\cdot \overline{YP}}{\overline{XP}\cdot \overline{YQ}}\right)$where P and Q are ideal points lying on the boundary of the unit disc, and $\overline{XQ}$ denotes the standard Euclidean distance between a point X inside the unit disc and an ideal point Q.Given the ideal points $P=\left(0,1\right),Q=\left(0,-1\right)$ , nad points $A=\left(0,0\right)$ and $B=\left(0,\frac{1}{2}\right)$ . I am asked to find the midpoint $M=\left(0,m\right)$ between A and B.

Feinsn 2022-06-16