4
votes
Accepted
Finf the segment EC in the triangle ABC below
Reflect $A$ over $BP$ to $A' \in BC$. Notice two things:
$|PA'|=|PA|=|PC|$, so $\triangle PA'C$ is isosceles with $|PA'|=|PC|$, thus $|A'E|=|EC|$.
$|BA'| = |BA| = 6$
To conclude, the remaining $|A'C|...
- 1,275
4
votes
Accepted
Find the angle $x$ in the triangle $APQ $ below
As in the figure below, let $R$ be the third vertex of the equilateral triangle $AQR$ (with $R$ on the same side as $B$, with respect to $AC$).
$ARB$ is isosceles by construction and the hypotheses, ...
- 7,613
3
votes
Accepted
Find the segment "DC" in the obtuse triangle below
Let $\angle BAD = \angle CAD = \alpha$ then $\angle BAD =\angle BCA = 2\alpha$
Draw $DE$ such that $AE=QE$ ($E$ is the midpoint of $AQ$)
then $AE=QE=DE=5$ and $\angle ADE = \angle DAE = \alpha$
then $\...
- 7,518
3
votes
Find the segment "DC" in the obtuse triangle below
Let $R$ be the midpoint of $AQ$, so $\lvert AR\rvert = \lvert RQ\rvert = 5$. Since $\triangle ADQ$ is right-angled, then $AQ$ is the diameter of its circumcircle, so $R$ is its center, which means ...
- 49.7k
3
votes
Accepted
Tetrahedron analogue of a triangle Cevians property
First, a proof for equation $(1)$ will be given. Then, an analogous method will be employed for the case of a tetrahedron.
Part 1: Proof for $(1)$
For a triangle $\triangle XYZ$, let $[XYZ]$ denote ...
- 1,275
3
votes
Find the angle $x$ in the triangle $APQ $ below
You've made a good start. I assume the angles in your diagram were provided with the problem. Thus, as you've determined, $\measuredangle ABP = 120^{\circ} - 2\alpha$ and $\measuredangle BPA = 60^{\...
- 49.7k
2
votes
Alternative proof of Apollonius's identity
If you want to proceed with Euclidean geometry then after proving the Pythagoras theorem (via rearrangement, similarity of triangles, etc); you can use it to prove Apollonius’s theorem.
Consider the ...
- 3,310
2
votes
How to prove opposite angle bisector theorem for convex quadrilaterals?
Regardless of where the bisectors of $\widehat B$ and $\widehat D$ intersect, let them meet $AC$ at $E$ and $F$ respectively. The angle you are looking for is $|\widehat{LEA} - \widehat{LFA}|$. We ...
- 1,975
2
votes
Accepted
How to prove opposite angle bisector theorem for convex quadrilaterals?
Let point $E$ on $AD$ intersection of $AD$ and extension of $BL$ .
$\angle BED = \angle A + \frac{\angle B}{2}$(exterior $\angle$ of $\triangle$)
$\alpha + \angle A + \frac{\angle B}{2} + \frac{\...
- 7,518
2
votes
Accepted
How to express an angle between two angle bisectors in interior angles of a convex quadrilateral?
$\angle LEC = \angle BEA = 180^{\circ} - \frac{\angle A}{2} - \angle B$
$\alpha + 180^{\circ} - \frac{\angle A}{2} - \angle B + \frac{180^{\circ} -\angle C}{2} = 180^{\circ}$ (in $\triangle LEC$)
$\...
- 7,518
2
votes
Accepted
Helly's theorem for $n\geq d+3$
I'm sure there's a neater way to do this, but:
Take any collection of $d+2$ of the $n$ convex set. By your proof they have a common intersection. Now if we multiply all $n$ sets by $\mathbb{R}$ to ...
- 5,633
2
votes
Accepted
Calculate the angle $x^o $ in the triangle below
The answer is $\boxed{\angle AFE = 25^\circ}$.
Two outer angle bisectors and one inner angle bisector are concurrent. In this case, for the triangle $\triangle ABE$, outer angle bisectors of vertex $B$...
- 1,275
1
vote
Find the segment "DC" in the obtuse triangle below
Using your result of $\measuredangle CQD = 90^{\circ} + \theta$ and the Law of sines with $\triangle CDQ$ gives
$$\begin{equation}\begin{aligned}
\frac{\lvert DC\rvert}{\sin(90^{\circ} + \theta)} &...
- 49.7k
1
vote
Calculate the angle $x$ in the quadrilateral $ABCD$
In the figure circles s and d have common chord CD so OF is perpendicular bisector of CD. We have:
$\overset {\huge\frown}{DH}=\overset {\huge\frown}{CN}$
$\Rightarrow \overset {\huge\frown}{NH}=\...
- 11.2k
1
vote
Find the measure of $\measuredangle BAC$ in the triangle below given two sides and an angle
Do you realise your angle might not be unique?
I've drawn a line segment with a length of $10$, an angle of $34.5°$ on it, and then I drew a circle with a radius of $6$, and this is what I came up ...
- 2,753
1
vote
Find the measure of $\measuredangle BAC$ in the triangle below given two sides and an angle
I would use the "sine rule:
$\frac{sin(34.5)}{6}= \frac{sin(A)}{10}=
\frac{sin(B)}{b}$.
$sin(A)= (10/6) sin(34.5)= (5/3)(0.5664)= 0.994$
$A= arcsin(O.994)= 70.7$ degrees.
70.7=
- 1,513
1
vote
How to prove opposite angle bisector theorem for convex quadrilaterals?
In your first figure:
$$
\begin{align}
\alpha &=\pi-\angle BLD\\
&=\pi-(2\pi-\angle B/2-\angle D/2-\angle C)\\
&={\angle B+\angle D\over2}+\angle C-\pi\\
&={2\pi-\angle A-\angle C\...
- 52.3k
1
vote
Area of the parallelogram formed by joining the midpoints of the sides of a quadrilateral
I'm afraid there is not a lot space for improvisation here. I suppose every proof of this fact will eventually end up chasing the middlesegment argument or more generally Thales's interception theorem....
- 507
1
vote
Area of the parallelogram formed by joining the midpoints of the sides of a quadrilateral
Another approach is to identify the intersection point of the two diagonals ($I$ on my diagram) and then observe that $S_{\triangle AEH} = S_{\triangle IEH}$ and so on.
To convince yourself that the ...
- 347
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