Problem: Let $ABC$ be a triangle and let its incircle touch $BC$ at $D$. Let $T$ be the   tangency point of the $A$-mixtilinear incircle with circumcircle of $ABC$. Prove that $∠BTA = ∠CTD$.   I recognised that it suffices to show that $KT$ is isogonal to $DT$.Then the first thing that came to my mind is harmonic quadrilateral.I don't whether it is true or not.I am new to the topic of isogonals.  Help me how to start??  Edit:I asked to prove that the above marked angles are equal.The question which marked as possible duplicate is not the same.Angles which were to be proven are different.I hope this convince you this as different question.I use mobile so geogebra in android is very difficult so I drew diagram manually.
123762
Let $FGH$ be a triangle with circumcircle $A$ and incircle $B$, the latter with touchpoint $J$ in side $GH$. Let $C$ be a circle tangent to sides $FG$ and $FH$ and to $A$, and let $D$ be the point where $C$ and $A$ touch, as shown here.    Prove that $\angle FGH = \angle GDJ$.
89213
NOTE: I would appreciate it if you provided a hint and not the whole solution.  BdMO 2014 Nationals:     In acute angled triangle ABC, considering a portion of side BC as diameter a    circle is drawn whose radius is 18 units and it touches AB and AC side. Similarly,    considering a portion of sides AC and AB as diameters, two other circles are    drawn whose radii are 6 and 9 units respectively. What is the radius of the incircle    of ∆ABC?   The main problem with this  problem is that the figure is incredibly messy.So I drew one circle,and tried getting information from it.I noticed that if $O$ is a centre of one of the above (semi)circles,and if $O$ is located on $AB$, then $CO$ is an angle bisector.Therefore,if we connect all the vertices with the centres,their intersection would be the incentre.Then we need to drop a perpendicular from this point to get the inradius.But that hardly helps us.I have found some similar triangles in the figure.But that does not help me at all.  I have also tried backtracking.I tried to put myself in the problem-maker's shoes and tried to imagine what I would do if I had to create a problem like this.Unfortunately,I couldn't do it..  Any insightful comment,hint will be very much  appreciated.
72286
Consider triangle ABC. Point D is on AC, E is on BC and F is on AB. Given that triangle DEF is equilateral and that segments CD, BE and AF are equal in length, prove that triangle ABC must also be equilateral. I've played with this one for long enough using circles, sine law, cosine law, etc., and have almost reached the point of capitulation. Any hints would be appreciated. I'm not actually sure if the statement is true, but I haven't been able to find a counterexample, either, so I'm stuck both ways.
98556
I want to prove, that $(\mathbb{R}^n,\|\cdot\|_{p})$ is isometrically isomorphic to $(\mathbb{R}^n,\|\cdot\|_{q})$ iff $p=q,$ I have tried looking the unitary balls, and I have been proved that $(\mathbb{R}^n,\|\cdot\|_{1})$ is not isometrically isometric to any other $p$-norm.
63211
I am pretty sure that this is such a stupid, stupid question, but how do you prove that $d_2(x,y)={|x-y|\over {1+|x-y|}}$ satisfying the third condition to be a metric, which is the triangle inequality.   I know that I should really be able to do this, but I just can't, I keep seeing a contradiction. Please help me.
49350
I tutor GCSE maths and I am stumped by this question:   Let XBYA be a quadrilateral The diagonals AB and XY intersect at point M  Given that the area of triangle AXB = the area of triangle AYB   Prove that XY is bisected by AB.  (I am guessing that students are expected to use the area of a triangle = 1/2 ab sin(C) but I cannot show XM=MY? )
119939
Let $ABC$ be a non-isosceles triangle. Medians of $\triangle ABC$ intersect the circumcircle in points $L,M,N$. If $L$ lies on the median of $BC$ and $LM=LN$, then prove that $2a^2=b^2+c^2$.  My Attempt:    Let $G$ be the centroid of $\triangle ABC$ and $D$ be the mid-point of $BC$.  Since $LM=LN$, therefore $LM$ and $LN$ will subtend equal angles on the circumference of circumcircle  $\Rightarrow \angle GBL=\angle LCG$  $GL=GL$  Altitude of $\triangle BGL$ from $B$ to $GL$=Altitude of $\triangle LGC$ from $C$ to $GL$  Area($\triangle BGL$)=Area($\triangle LGC$)  Now, based on this ,can it be said that $\square GBLC$ is a parallelogram. If yes then   $$DL=GD=\frac{m_{a}}{3}$$  $\Rightarrow AD.DG=BD.DC$  $$m_{a}.\frac{m_{a}}{3}=\frac{a^2}{4}$$  $\Rightarrow 4m^2_{a}=3a^2$  $\Rightarrow 2b^2+2c^2-a^2=3a^2$  $\Rightarrow b^2+c^2=2a^2$  I am not sure whether this justification is sufficient(the one that I have written in bold). What more can be added to seal the issue
43128
We have given isosceles triangle $ABC$ with baseline $AB$. Point $M$ is midpoint of AB. Draw perpendicular line to side $AC$ through $M$ which intersects $AC$ in point $H$. Let $P$ be midpoint of $MH$. Show that lines $BH$ and $CP$ are perpendicular to each other.  I also have "hint": Construct height $BD$ to side $AC$ and look at the triangles $ABD$ and $MCH$.  Analytically there is no problem but I get stuck in geometrical approach to solution.
118076
please help with this proof.  "Show that the tangents at the endpoints of a focal chord of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ meet on the corresponding directrix."  This is a homework question with two part where the first part is to prove the converse of the above statement (namely prove that the chord of contact from a point on directrix is a focal chord)  For the first part I was able to utilize the following equation for chord of contact from a point $P(x_0, y_0)$  \begin{align} \frac{x_0 x}{a^2} - \frac{y_0 y}{b^2} = 1 \end{align}  Then substitute $x_0$ for $\pm \frac{a}{e} $ (x coordinate of directrix) and then show that $S(\pm ae, 0) $ (coordinate of focuses) satisfy the equation.  But it seem I could not use the same method to show the second part.  Please help. Thanks.
36735
Here there is my proof (quite short and easy) of a rather straightforward result. The text of this question comes from a previous question of mine, where I ended up working on a wrong inequality. Here things are fixed. Still, I would like to know:  if it is sound, because absolute value always creates me some problem, and if there is a shorter (neater) way to get the result (maybe without using contradiction).    Proposition: $|d(x,y)-d(y,z)| \leq d(x,z)$. Proof: We proceed by cases.  Case 1: $d(x,y)-d(y,z) \geq 0$. Assume by contradiction that $d(x,y)-d(y,z) &gt; d(x,z)$. Hence, $d(x,y)&gt; d(y,z) + d(x,z)$, contradicting the triangle inequality.  Case 2: $d(x,y)-d(y,z) &lt; 0$. Assume by contradiction that $ - d(x,y) + d(y,z) &gt; d(x,z)$. Hence, $d(x,y) &lt; d(y,z) - d(x,z)$. By applying the triangle inequality, we have that $d(y,x) \geq d(y,z) + d(x,z)$, which is by symmetry equivalent to  $d(x,y) \geq d(y,z) + d(x,z)$, obtaining the desired contradiction. $\square$     As always, any feedback is welcome. Thank you for your time.
117125
Triangle $\triangle ABC$ is a right triangle with $\measuredangle ACB=90^\circ$. Let $AB=c$ and the radius of the inscribed circle be $r$. Find the catheti and the area of the triangle $\triangle ABC$.    Let $P,N$ and $M$ be the tangent points with $AB,BC$ and $CA$, respectively. The quadrilateral $MINC$ is a square. Therefore, $MI=IN=NC=CM=r.$ We have $AM=AP=c_1$ and $BN=BP=c_2$ as tangent segments. Now we have: $$\begin{cases} (c_1+r)^2+(c_2+r)^2=(c_1+c_2)^2 \\ c_1+c_2=c \Rightarrow c_2=c-c_1 \end{cases}.$$ After simplifying the first equation I got $r^2+c_1r+c_2r=c_1c_2$ and now let us plug $c_2=c-c_1$. We get the quadratic equation $c_1^2-cc_1+r^2+cr=0$ which has roots $c_1=\dfrac{c\pm\sqrt{c^2-4r^2-4cr}}{2}$. I am not sure I understand what to do next and why we got two expressions for the same segment. What does that mean? If use use both the results in $c_2=c-c_1=\dfrac{c\pm\sqrt{c^2-4r^2-4cr}}{2}$. What does mean?
103263
Let ABCD be a cyclic quadrilateral. The diagonals AC and BD meet at P, and DA and CB produced meet at Q. The midpoint of AB is E. Prove that if PQ is perpendicular to AC, then PE is perpendicular to BC.
29160
Let $PQRS$ be a trapezoid. The bisectors of legs $PS$ and $QR$ intersect the opposite legs at $M$ and $N$ so that triangles $PMS$ and $QNR$ are formed.  Prove that $\measuredangle PMS= \measuredangle QNR$.  The question may seem easy but I fail to move on, playing with angles didn't help. I'd appreciate some hint from somebody with an eye for geometry and triangles.
92237
Hi all, I am interested to find elementary proof of tangent half angle formula.  My solutions are the following:  Triangle $AOB$ is such that $|AB|=1$ and $\angle AOB=\theta$. We then extend $OB$ to $P$ and $Q$ such that $|OP|=|OQ|=1$. Thus we will have two isosceles triangles: $AOP$ and $AOQ$.  From the picture, $\tan{\left(\frac{\theta}{2}\right)}=\frac{AB}{BP}=\frac{\sin{\left(\theta\right)}}{1+\cos{\left(\theta\right)}}\ \ \ =\frac{BQ}{AB}=\frac{1-\cos{\left(\theta\right)}}{\sin{\left(\theta\right)}}$  Could You guys please check my solution. I am also wondering if there are other elementary solutions, please share, thanks!
11965
struggling with a question from homework and would appreciate some assistance.  Let $A, B \in M_2^{\mathbb{R}}$ be defined as follows: $$A = \begin{pmatrix} 2 &amp; 1 \\ 1 &amp; 1 \end{pmatrix}, B = \begin{pmatrix} -4 &amp; 1 \\ 1 &amp; 1 \end{pmatrix} $$ Find a regular $P \in M_2^\mathbb{R}$ such that $P^tAP=I$ and $P^tBP$ is diagonal.  I'm familiar with the theorem that states that a simultaneous diagonlization exists if one of the matrices is positive-definite (or negative-definite). I've found an invertible P such that $P^tAP=I$: $$P=\begin{pmatrix}1\over\sqrt{2}&amp;-1\over\sqrt{2}\\0&amp;\sqrt{2}\end{pmatrix}$$ But $P^tBP$ is not diagonal. Would appreciate any assistance / hints. Thanks!
68989
The question is as follows:      Let $A'$, $B'$ and $C'$ be the midpoint triangle of triangle $ABC$. In other words, $A'$, $B'$ and $C'$ are the midpoints of segments $BC$, $CA$ and $AB$, respectively. Show that triangles $A'B'C'$ and $ABC$ have the same centroid.    My reasoning was through the means of the Midline Theorem. After drawing the diagram, we can see, for example, $C'A'$ is a midline of $\triangle ABC$, which means that $C'A'$ is parallel to the base $AC$ and half the length of it as well. Therefore, the median $B'B$ would pass through the midpoint of $C'A'$, including the midpoint of $AC$.   Is my reasoning valid or, at least, is it on the right track? Any advice on solving this problem will be greatly appreciated.
121601
This question arose while I was answering this question, (). First the original question:  $ABCD$ is a quadrilateral. A line through $D$ parallel to $AC$ meets $BC$ produced at $P$ we need to show $$Area(\Delta APD)=Area(ABCD)$$    Its easy to see that the OP must have meant to prove, $Area(\Delta ABP)=Area(ABCD)$, the proof of which is given in my answer. However, it set me thinking when actually $Area(\Delta ADP)=Area(ABCD)$ is true? I have not been able to derive any good conclusion (I mean, in terms of the elements (diagonals, angles or sides) of $ABCD$). Can anyone help?