Here is a problem from a sample paper:  I see that $\rm area(ACE) = area(DCE)$.  I also see that $\rm AB$ is the common base for $\triangle \rm ABE$ and $\rm ABCD$.  So this means that the height of $\rm ABCD$, i.e., $\rm AD$ is exactly the half of $\rm BE$.  I can't seem to get around the rest, though.   It seems that $\rm ACED$ is a parallelogram, but that'd mean $\rm BC = CE \Rightarrow AD = DC$ which doesn't seem so looking at the figure.  Could anyone please give me a complete proof using only the theorems:   Two quadrilaterals lying on the same base and between the same parallels have the same area. Two triangles lying on the same base and between the same parallels have the same area.   I also know that the area of 1 is twice the area of 2.
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?
107441
I had some troubles with this problem :     Let $ABCD$ be a convex quadrilateral. $M$ and $N$ are the midpoints of   the diagonals $AC$ and $BD$. The sides $AB$ and $CD$ are extended   until they intersect. The intersection point is $E$. The sides $AD$   and $BC$ are extended until they intersect. The intersection point is   $F$. Let $P$ be the midpoint of the segment $[EF]$. Prove that $M$, $N$, $P$ are    collinear.     First, I found that a quadrilateral in which the opposite sides interesect is also known as a complete quadrilateral. Then, the line $M-N-P$ is known as Newton-Gauss line and the problem above as Newton's Problem.  I've taught about solving it using areas. I've used the property that median divides the triangle in two echivalent triangles (with the same area). Many properties can be derived from it.  I haven't figure out, but I'm interested in a proof using areas. I would appreciate some suggestions.  Thanks!
105893
A student sent me a question and I am stucked. I'd like to use classical geometry, instead of hard trigonometry if possible (can be trigonometric, but I'd like to not use scientific calculator; the only answer I got I need this), but I am in circles. The pentagon $ABCDE$ is regular. The answer is $48^\circ$ (I've constructed in Geogebra). Thank you in advance!
118281
[Update: Based on the hints provided by @zcn and @whacka, I believe I have found a solution. See my answer below.]  Below, $R$ is a commutative ring with $1$.  In John J. Watkins' Topics in Commutative Ring Theory, the author observes that if $a$ and $b$ are associates in $R$, then they generate the same ideal: $(a) = (b)$. This is clearly true, because $a = ub$ and $b = va$ for some units $u$ and $v$, and so $a \in (b)$ and $b \in (a)$, whence $(a) \subset (b)$ and $(b) \subset (a)$. Indeed, we didn't even require $u$ and $v$ to be units to reach this conclusion.  The author then claims that if $R$ is not a domain, then the converse may not be true: we may have $(a) = (b)$ even if $a$ and $b$ are not associates. As a "counterexample" he gives $R = \mathbb{Z}/(6)$, with $a = 2 + (6)$ and $b = 4 + (6)$. Clearly these elements generate the same ideal: $(a) = (b) = \{0, 2 + (6), 4 + (6)\}$. The author asserts that $a$ and $b$ are not associates, but this is false: $5 + (6)$ is a unit, and $(2 + (6))(5 + (6)) = 4 + (6)$, so $a$ and $b$ actually are associates.  I figured that the author botched the example but that surely the basic fact must be true: if $R = \mathbb{Z}/(n)$ then there may be nonassociates $a,b \in R$ with $(a) = (b)$. However, I was not able to find a counterexample and eventually resorted to writing a computer program and found that there are no counterexamples in $\mathbb{Z}/(n)$ for $n \leq 100$.  So now I'm starting to think that there is no counterexample in $\mathbb{Z}/(n)$ at all, but I have not yet been able to prove this.  My questions:   If I am correct that $(a) = (b)$ in $\mathbb{Z}/(n)$ implies that $a$ and $b$ are associates, I would appreciate a nudge in the right direction toward a proof. (But if the proof is elementary, please give only a hint, not a solution :-) I'll post my solution once I have one. In either case, could you please point me to a genuine counterexample? If there is one in $\mathbb{Z}/(n)$ for some $n$, that would be ideal (no pun intended), otherwise a counterexample in any commutative ring with $1$ would be great.
103423
I read this recently on the web and can't manage to understand it. Not homework -- I haven't done math homework for years.  If $d\mid ab$ then $d=d_1 d_2$, that $d_1|a$, that $d_2|b$ and, further, if $(a,b)=1$ then $(d_1,d_2)=1$.  I know it's the same question... if $a = a_1 a_2 ... a_n$, why can't $d$ divide some $a_i$ and only $a_i$?
123785
Let $ABC$ and $PQR$ be any two triangles in the same plane. Assume that the perpendiculars from the points $A,B,C$ to the sides $QR,RP,PQ$, respectively, are concurrent. Prove that the perpendiculars from $P,Q,R$ to $BC,CA,AB$ respectively are also concurrent.   I was suggested to prove this using vectors. But I have no idea how to use them...
63019
I found this question in an exam question paper.  But I was stumped by it. It seemed that all the answers were correct.  I tried:    Let $\angle ADE=x$ and $\angle EDC=y$.  $\angle DAB=90^{\circ}=\angle DAE+\angle EAB=30^{\circ}+60^{\circ}$. $\text{ref}\angle AEB=300^{\circ}$. And we know that $x+y=90^{\circ}$  But now I am stuck. How to proceed?
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? )
122003
See below in acute triangle $ABC$, $D$ is on $AC$ such that $AD=BC$. $CF$ is the angle bisector of $\angle ACB$. $DE \parallel CF$. $E$ is on $AB$. $AE = CD$.  Prove that $\angle ABC = 2 \angle BAC$. In addition, find the condition in which $\angle ADB = 3 \angle BAC$  I struggle to find ways to use two conditions $AD = BC$ and $AE = CD$
61457
The question is as follows if $ABC$ is a triangle, with $AD$ as the internal angle bisector of $\angle A$ with $D$ at $BC$ and $B^{'}, C^{'}$ are reflection of points $B$ and $C$ in $AD$. Show that triangles $ABC$ and $AB^{'}C^{'}$ have the same incentre. This was the exact question language,     Now intuitively here was where I could not satisfy myself, how come two triangles which only shared a vertex $A$ above have the same incentre ?all I could find in this problem was we have $C^{'}B^{'} = CB$,(because notice that here $B$ and $B^{'}$ lie at a same distance from $AD$ as $B^{'}$  is a reflection of $B$ only,same with $C$ and $C^{'}$) which also gives us $BD = DB^{'}$ and $CD = DC^{'}$ , all I claim here could be proved if we join $BB^{'}$ and $CC^{'}$ which pass through $AD$ and is perpendicular to it and also these 2 pair of points($B, B^{'}$) &amp; ($C, C^{'}$) lie equidistantly from $AD$, by which I could say triangle $BDC^{'}$ is congruent to $B^{'}DC$. I didn't find anything more substantial.  So here is my question from the figure can we ever have the incentre of triangle $ABC$ to be the incentre of  $AB^{'}C^{'}$ for incentre we need a centre of a circle inside the triangle which touches all the 3 sides of it which so how come another circle of a different triangle have the same incentre ? Although I could not disprove it concretely it just runs counterintuitive to me, so do I miss something Or did I interpret the question wrongly ?
73885
Can anyone give me a vectorial solution to the following problem:     Prove that if each pair of opposite edges of the tetrahedron $ABCD$ is perpendicular (that is, $AB \perp CD$ and $AC \perp BD$ and $AD \perp BC$), then the heights of the tetrahedron are concurrent.   Here, the heights (also known as the altitudes) of tetrahedron $ABCD$ are the perpendicular from $A$ to the plane $BCD$, and three other similarly defined perpendiculars.  Tetrahedra satisfying the condition of this problem are called , and this appears to be a known result.
23698
I was looking for proofs using Calculus for the area of a circle and come across this one $$\int 2 \pi r \, dr = 2\pi \frac {r^2}{2} = \pi r^2$$  and it struck me as being particularly easy. The only other proof I've seen was by a teacher and it involved integrating $x = \sqrt{r^2 - y^2}$ from $-1$ to $1$, using trig substitutions and then doubling the area to get $\pi r^2$ but the above proof seemed much more straight forward.  Is it a valid proof, or is it based on circular logic or some other kind of fallacy?
22566
I have some troubles with this problem :     Let $ABCD$ be a convex quadrilateral. $M$, $N$, $P$ and $Q$ are the   midpoints of the sides $AB$, $BC$, $CD$ and $AD$.  $AN$, $BP$, $MD$   and $CQ$ are interescting in $X$, $Y$, $Z$ and $T$ like in the figure   below. Prove that $[XYZT] = [AMX] + [BYN] + [CZP] + [DTQ]$.   It is noted with $[ABC]$ the area of $ΔABC$.     Since $M$, $N$, $P$ and $Q$ are midpoints, the first thing that came in my mind was the median property : the median divides a triangle in two echivalent triangles (with the same area).  I would appreciate some suggestions.  Thanks!
106327
Could you help me?  Find the area ratio between ABC and the hatched triangle.    Thanks in advance.
84719
Question :  Let $z_1,z_2,z_3,z_4$ be the position vectors of the vertices for quadrilateral $ABCD$. Prove that $ABCD$ is a parallelogram if only if $z_1-z_2-z_3+z_4=0$  What i've done so far :  Proof of $p\Rightarrow q$  Let $z_1=x_1+iy_1,\quad z_2=x_2+iy_2, \quad z_3=x_3+iy_3, \quad z_4=x_4+iy_4$  $ABCD$ is a parallelogram $\Leftrightarrow\,\overrightarrow{AB}=\overrightarrow{CD}\, \land\, \overrightarrow{AD}=\overrightarrow{BC} $   $\begin{aligned} \overrightarrow{AB}=\overrightarrow{CD} &amp;\Rightarrow |z_1-z_2|=|z_3-z_4|\\ &amp;\Leftrightarrow x_1^2-2x_1x_2+x_2^2+y_1^2-2y_1y_2+y_2^2=x_3^2-2x_3x_4+x_4^2+y_3^2-2y_3y_4+y_4^2\\ \overrightarrow{AD}=\overrightarrow{BC} &amp;\Rightarrow |z_1-z_4|=|z_2-z_3|\\ &amp;\Leftrightarrow x_1^2-2x_1x_4+x_4^2+y_1^2-2y_1y_4+y_4^2=x_2^2-2x_2x_3+x_3^2+y_2^2-2y_2y_3+y_3^2 \end{aligned}$  I'm stuck and see no conclusion here.  Proof $p\Leftarrow q$  I don't have an idea for this.  Please help me, thanks.
61878
Let $ABCD$ be a convex quadrilateral such that the length of the segment connecting midpoints of the two opposite sides $AB$ and $CD$ equals $\frac{AD+BC}{2}$ . Prove that AD is parallel to BC. I assume that AD is not parallel to BC but I can't find any contradiction.
13874
If $\triangle$ be the area of $\triangle ABC$ with side lengths $a,b,c$. Then show that $\displaystyle \triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$   and also show that equality hold if $a=b=c$.  $\bf{My\; Try}::$ Here we have to prove $4\triangle\leq \sqrt{(a+b+c)\cdot abc}$  Using the formula $$\triangle = \sqrt{s(s-a)(s-b)(s-c)},$$ where $$2s=(a+b+c)$$  So $$4\triangle = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{(a+b+c)\cdot(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$  Now using $\bf{A.M\geq G.M}$ for $(b+c-a)\;,(c+a-b)\;,(a+b-c)&gt;0$  $$\displaystyle \frac{(b+c-a)+(c+a-b)+(a+b-c)}{3}\geq \sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$  So we get $\displaystyle (a+b+c)\geq 3\sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$  But I did not understand how can I prove above inequality  help Required  Thanks