Acute $\triangle{ABC}$, point $D$ is on side $AC$ so that $AD=BC$, $CF$ bisects $\angle{ACB}$. Point $E$ on side $AB$ so that $DE \parallel CF$ and $AE=CD$. Prove: $\angle{ADB}=3\angle{BAC}$
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 $AE = CD$
117366
Let $ABC$ be an acute-angled triangle. The feet of the altitudes from $A,B$ and $C$ are $D, E$ and $F$ respectively. Prove that $DE +DF \le BC$ and determine the triangles for which equality holds. The altitude from A is the line through A which is perpendicular to BC. The foot of this altitude is the point D where it meets BC. The other altitudes are similarly defined.   Thanks in advance for any contributions.
121292
Points $P$, $Q$, $R$ and $S$ are chosen on the sides of parallelogram $ABCD$, so that $P$ is on line $AB$, $Q$ is on line $BC$, $R$ is on line $CD$, $S$ is on line $DA$, and $AP=BQ=CR=DS=\frac{1}{3}(AB)$. Compute the ratio of the area of $PQRS$ to the area of $ABCD$.
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.
61841
I need a synthetic proof on this problem without the use of trigonometry.     Question: Let $ABC$ be a triangle with $AB=AC$. If $D$ is the midpoint of $BC$, $E$ the foot of perpendicular drawn from $D$ to $AC$ and $F$ the midpoint of $DE$, prove that $AF\perp BE$.    The proof is very easy using coordinate geometry.    I'm stuck at: Since point $F$ is midpoint of $DE$, it is fixed and is not a trivial information. But I can't see a way to use this information.
54287
Let a, b, c be the three side lengths of a triangle. Prove that  $$\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\geq 3$$  Under what conditions is equality obtained?
34932
I need some help proving this, I've seen it proven in the other direction (prove the formula if it is an equilateral) but cant figure out how to prove it this way around.     Given three complex numbers $z_1, z_2, z_3$ prove that the points $z_1, z_2, z_3$ are vertices of an equilateral triangle in $\Bbb C$, if $$z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_1z_3 + z_2z_3$$
41974
To prove that if $ab \equiv ac\bmod n$ and $(a,n)=1$ then $b\equiv c \bmod n$  I write as  $1=ar+ns$  $ac=ab+nq$  I have to prove $c=b+nr$  How do I manipulate equations to reach conclusion  Thanks
109975
Let $a&gt;0$,$b&gt;0$ and $c&gt;0$. Prove that:    $$\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le \dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.$$     $$\Leftrightarrow \frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+b+c)}{b+c}+\frac{(c^2+a^2)(a+b+c)}{c+a}\leq 3(a^2+b^2+c^2)$$  $\Leftrightarrow 2(a^2+b^2+c^2)+\frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq 3(a^2+b^2+c^2)$  $\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq a^2+b^2+c^2$  $\Leftrightarrow \frac{c(a+b)^2-2abc}{a+b}+\frac{a(b+c)^2-2abc}{b+c}+\frac{b(a+c)^2-2abc}{a+c}\leq a^2+b^2+c^2$  $\Leftrightarrow 2(ab+bc+ac)\leq a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)$  I can't continue. Help me
87442
For an arbitrary convex quadrialteral $ABCD$:   $$AF=FC, BE=ED, GI=IH$$  Prove that points $F$, $E$ and $I$ are collinear.  I was able to solve the problem by using vectors and I think that it can be also solved analytically but I wanted some more elegant proof, more in the spirit of Euclid.
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.
8018
I just need to check the reasoning in my proof is correct, I think it is valid although I'm not totally convinced because I can't follow the logic; does proving that $x$ is an integer prove that $ac|bd$?  Theorem: Let $a$, $b$, $c$, $d$ be integers. If $a|b$ and $c|d$, then $ac|bd$.  Proof: $aj=b$ and $ck=d$ for integers $j$, $k$. Then, $ac|bd$ implies $acx=ajck$ and thus $x=jk$ for some $x$. Since the product $jk$ is an integer, $x$ is an integer and thus $acx=ajck$ and thus $ac|bd$.
96035
Convex hexagon $ABCDEF$ inscribed within a circle. Prove that $$AC \cdot BD \cdot CE \cdot DF \cdot AE \cdot BF \geq 27 AB \cdot BC \cdot CD \cdot DE \cdot EF \cdot FA\,.$$  I was thinking of represending the inequalities in trigonometry then use Language multiplier. For example let $\angle AOB = \theta_1$, $\angle BOC = \theta_2$, represent the inequality in trigonometry, subject to constraint $\theta_1 + \theta_2 + ... + \theta_6 = 2\pi$. But it's still quite a bit of work and I didn't manage to get to the end. It also seems a bit overkill -- might be better solution? Would like to see any approach.
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.
16982
This question looks really simple, but to my (and my co-workers) frustration we were not able to prove this in any way. I tried all triangle formulas known to me but I feel I'm missing the point, and proof will be or much simpler or much more complicated than what I tried.  So, the question:  Given a triangle ABC and point P inside that triangle, prove that for triangle APB the following inequality holds: |AB| + |BC| > |AP| + |PC|  (Actually it doesn't matter for me if it's > or >=).
21150
Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$  A full expanding results in: $$\sum_{cyc}a(a+b-c)(a+c-b)\geq3\prod_{cyc}(a+b-c),$$ or $$\sum_{cyc}(a^3-ab^2-ac^2+2abc)\geq\sum_{cyc}(-3a^3+3a^2b+3a^2c-2abc),$$ but it becomes very ugly.
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!