Right triangle and two parallelograms

Let's assume AB=a, BC=b and CA=c. X is the point so that VX is parallel to AB and UX is parallel to BC. So VXU is a right triangle at X. Let r be the inradius of ABC, so r = (a+b-c)/2. Let's calculate VX and UX.

VX = (1/2)IP + r - (1/2)AB = (1/2) (AM + 2r - a) = (1/2) (AM + (b-c))

By the angle bisector theorem AM/MB = c/b, so AM/a=c/(b+c), or AM = ac/(b+c) . Hence

(b+c)(AM + (b-c)) = ac + (b^2-c^2) = ac - a^2 = a(c-a) = a(c-a)(c+a)/(a+c) = ab^2/(a+c)

So 2(a+c)(b+c)VX = ab^2 Similarly 2(a+c)(b+c)UX = ba^2

So VX/UX= b/a = CB/AB. Hence VXU is similar to CBA. This should be enough to conclude UV is perpendicular to AC.

Wow I'm so happy that the new features have given us not only geometry problems, but also more discussions.

This is a very nice problem here I will give a synthetic solution:

Let's take advantage of all these midpoints. We construct a point on ray$AV$ such that $V$ is the midpoint of $AK$.Now we have $VU\parallel CK$, which means we just have to prove $CK\perp AC$. Moreover, we get parallelogram $APKQ$ so $QK=MI, IN=QC, \angle KQC=\angle CIN=45$. Now since $\angle QCA=\angle CAN=\angle NAB$, therefore $CK\perp AC$ is equivalent to proving $\angle KCQ=90-\angle QCA=90-\angle NAB=\angle ANB$. Since $\angle KQC=45=\angle NBI$, thus we just have to prove $\triangle KQC\sim \triangle IBN$ or simply $\frac {BI}{BN}=\frac {QK}{QC}=\frac {MI}{IN}$. This is indeed true from $\triangle MIB\sim \triangle INB$ and here's its proof: since $\angle AIM=\angle ABI=45$, therefore $\angle NIB=\angle IAM+\angle IBA=\angle IAM+\angle AIM=\angle IMB$. It's already known that $\angle IBM=\angle IBN$, so that gives us the similarity. $\Box$

I haven't been keeping up with all these posts lately, wish I had found this sooner. :)

The following is a nice property about a right triangle.

[Keeping the same notation of our problem]

Property
The projections of the segments $MI$ and $IN$ on the line $AC$ have the same length.

Proof Let $E$ and $F$ in $AC$ such that $ME$ and $NF$ are perpendicular to $AC$. Denoting $d(X,YZ)$ the distance from point $X$ to line $YZ$, we have $d(I, ME)=d(I,MB)=r=d(I, NB)=d(I,NF)$ . Since $d(I,ME)$ is the length of the projection of $MI$ on the line $AC$ and $d(I,NF)$ is the length of the projection of $NI$ on the line (AC), we are done!

Could you use this property to solve our problem?

Are we allowed to use co-ordinate geometry? I thought of a solution using co-ordinate geometry but it is turning out to be a bit ugly.

Please post a solution that is understandable and graceful......... I'm still in Middle School

AC is parallel PQ UV &AC are perpendicular