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Nice geo
Let $P,Q$ be the feet of perpendiculars from the orthocentre $H$ on the internal and external angle bisectors of $A$ respectively of triangle $ABC$.Let $M$ be the midpoint of $BC$. Prove that $P,Q$ and $M$ are collinear.
- Mon Dec 12, 2016 12:25 am
- Forum: Combinatorics
- Topic: Forming Triangle
- Replies: 1
- Views: 2390
Re: Forming Triangle
fibonacci does the trick
- Mon Dec 12, 2016 12:24 am
- Forum: Combinatorics
- Topic: Divisible by 100
- Replies: 1
- Views: 2386
Re: Divisible by 100
Mod 100 and php
- Tue Nov 15, 2016 12:00 am
- Forum: Geometry
- Topic: Prove parallel
- Replies: 2
- Views: 3140
Re: Prove parallel
Solution: We reflect the whole figure except the triangle $ABC$ under the perpendicular bisector of $BC$. Let the Point $X'$ represent the reflection of a point $X$. Now, we can easily prove by various calculations that there exist a circle tangent to both $AB$ and $AC$ and passes through the points...
- Wed Oct 19, 2016 8:41 pm
- Forum: Geometry
- Topic: Tangent Circles
- Replies: 4
- Views: 4753
Re: Tangent Circles
$KPG ,KDM$asif e elahi wrote: Which two circles?
- Thu Oct 13, 2016 9:36 pm
- Forum: Geometry
- Topic: Tangent Circles
- Replies: 4
- Views: 4753
Re: Tangent Circles
$Hints:$ Prove a lemma: $W,W'$ are two circles which are reflections of eachother under $BC$. Let $M$ be the midpoint of $BC$. Let $P,Q$ be two lines intersecting $W,W'$ at respectively such that $U,V,W,X$ lies on the same side of $BC$. Then $UVWX$ is cyclic. Then, show that the centres of two respe...
- Wed Oct 12, 2016 9:07 pm
- Forum: Geometry
- Topic: Useful lemma
- Replies: 1
- Views: 3337
Useful lemma
In triangle $ABC$ the $incircle$ touches $BC$,$CA$ and $AB$ at $D$,$E$ and $F$ respectively. Let the mid-points of $BC$,$CA$ and $AB$ be $M$,$N$ and $P$ respectively. Prove that, $1)$ $AI$,$DE$,$MP$ and the perpendicular from $B$ to $AI$ are concurrent. $2)$ $AI$,$DF$,$MN$ and the perpendicular from...
- Mon Oct 10, 2016 11:52 am
- Forum: Geometry
- Topic: Perpendicular lines through the foot of an altitude
- Replies: 2
- Views: 4149
Re: Perpendicular lines through the foot of an altitude
Solution: Here, $\textit{Butterfly Theorem}$ gives us $DE=DF$. Again it is a well known fact that the reflection of $H$, be $G$ lies on the circumcircle of $ABC$. Now, in quadrilateral $HEGF$, the diagonals bisect each other. Therefore, it is a parallelogram. Now, $\angle EHD=\angle AGC=\angle ABC$ ...
- Tue Aug 16, 2016 8:13 pm
- Forum: Geometry
- Topic: A self-made geo
- Replies: 3
- Views: 3573
Re: A self-made geo
Solution:
- Mon Aug 15, 2016 8:40 pm
- Forum: Geometry
- Topic: Prove it's perpendicular
- Replies: 1
- Views: 2432
Prove it's perpendicular
Let one of the intersections of two circles with centres $O1$ and $O2$ be $P$. A common tangent touches the circles at $A$,$B$ respectively.Let perpendicular from $A$ to $BP$ meet $O1O2$ at $C$. Prove that $AP$ is perpendicular to $PC$.