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Results tagged with geometry
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user 783885
For questions about geometric shapes, congruences, similarities, transformations, as well as the properties of classes of figures, points, lines, and angles.
1
vote
Find the angle in the given quadrilateral
The best proof by far:
$B$ is clearly the $D$ excenter of $\triangle ADC$, therefore $DB$ is an angle bissector of that triangle.
So: $\angle CDB = \frac{180^{\circ} - 60^{\circ} - 40^{\circ}}2 = 40^{ …
- 2,080
-1
votes
Constructing a circle that internally tangents a circle $\gamma$ and passes through two inte...
I will give you the general solution of the apollonius PPC which works fine for this case:
draw any circle $\Gamma$ that passes through $A$ and $B$ and meets $\gamma$ in two points $M$ and $N$.
Let $ …
- 2,080
2
votes
Find ratio formed by the perpendicular bisector of an angle bisector
another approach. Let $X$ be the foot of the external angle bissector of $A$ in $\triangle ABC$ and without loss of generality, let's define $CD = 3 \implies BD = 4$. Now let $x = XC$, then: $$\frac{x …
- 2,080
0
votes
Internal Bisector of a triangle
$[ABC] = \frac{ab \sin c}2$. Let $D$ be the feet of the internal $A$ bisector.
$[ADB] + [ADC] = [ABC] \iff cd\sin(\frac{\angle A}2) + bd \sin(\frac{\angle A}2) = bc \sin(\angle A) \iff$
$ AD = \frac{2 …
- 2,080
4
votes
5
answers
295
views
Show that the bissectors of $\angle BHC$ and $\angle BFC$ meet on $BC$
Given an acute triangle $\triangle ABC$ with orthocenter $H$. Let $D = BH \cap AC, E = CH \cap AB$ and $F = (AEDH) \cap (ABC) \neq A$. Show that the inner angle bisectors of $\angle BFC$ and $\angle B …
- 2,080
1
vote
2
answers
95
views
Show that $\angle BFD = \angle CFD$
Given a $\triangle ABC$ whose incenter is $I$ and its incircle is $\gamma = \odot (I,ID)$ with $D \in BC$.
Also, define $\{E,D\} = \gamma \cap AD$ and $F$ as the midpoint of $ED$.
Show that $\angle BF …
- 2,080
2
votes
4
answers
128
views
Prove that $M$ is the midpoint of $HG$
In the image, $A,B$ and $E$ are tangency points;
$H,E,M$ and $G$ are collinear;
$O$ is the center of the circle whose diameter is $AB$
$OM \perp HG$
I did solve this problem with the following simila …
- 2,080
1
vote
Prove that $M$ is the midpoint of $HG$
Following Ivan Kaznacheyeu's comment:
Let $X = OG \cap AH$, then $\triangle XAO \cong GBO$ ($A-S-A$ on sides $OA=OB$).
So $OX=OG$
$\angle XHG = 90^{\circ} \implies HO=OG=OX \implies \triangle HMO \con …
- 2,080
3
votes
Accepted
Given a triangle ABC inscribed in the unit circle
line $AH$:
$$\frac{z-a}{b+c} = \overline{(\frac{{z-a}}{b+c})}$$
line $BC$:
$$\frac{z-b}{c-b} = \overline{(\frac{{z-b}}{c-b})}$$
so in the first equation:
$$\bar z = \overline{b+c} \cdot \frac{z-a}{b+c …
- 2,080
2
votes
3
answers
190
views
How can I prove a line can't meet an ellipse in more than two points
Without knowing the equation for an ellipse, but rather just with the geometric definition i.e. the ellipse is the locus of points with $PF_1 + PF_2 = k > F_1F_2$.
I tried to use a classic constructio …
- 2,080
1
vote
0
answers
75
views
How to prove that a circle inscribed in a sphere will remain a circle after a random movement
I'm trying to prove Euler's theorem for rigid bodys (in 3-D space of course).
I would like to prove the following: given a sphere $\mathcal S$ and circle $\gamma \subset \mathcal S$, assuming $\mathca …
- 2,080
1
vote
Center of wheel travels the length of circumference in one revolution
What you described is not a rule, it's only the case when you have rolling without slipping.
These two lengths you've described are usually independent on each other.
- 2,080
7
votes
2
answers
426
views
Show that $M$ is the midpoint of $JI$
Projective geometry also doesn't seem to help except if we could use Pascal's theorem in a tricky way, perhaps with the symedian of $\triangle ABC$ but I'm not sure how to do it. …
- 2,080
1
vote
solution-verification | Under what conditions is $AMC'N$ square?
Let's suppose that there is such $M$ in the middle of $BB'$ such that $AMC'N$ is a square.
It is well known that, in space, the locus of the points who see a segment $XY$ under a right angle, i.e. $\a …
- 2,080
2
votes
Accepted
Construct Triangle $ABC$ Given Intersections of Medians with Circumcircle
From this file we know that the centroid of $\triangle ABC$ is one of the two foci $G_1$ and $G_2$ of the Steiner inellipse of $\triangle DEF$. Therefore all you need is to find these points and draw …
- 2,080