# Cylic Trapezoid in a Triangle

In $\Delta ABC, P, Q$ and $R$ are the midpoints of $AC, AB$ and $BC$ respectively. Lines $PR$ and $QS$ intersect at $T$. If $BS\perp AC$ , prove that $TS=TP$ .

Solution

Connect points Q and R. Then by midpoint theorem, $AC\parallel QR$ and $2QR=AC$.

Let M be the intersection of SB and QR.

Since SB is the height of $\Delta ABC$ , then MB is the height of $\Delta QBR$ (i.e. $QR\perp MB$ ). It follows that M is the midpoint of SB and $\Delta SRM\cong \Delta BRM$ by SAS postulate. Hence, $SR=BR$ . As R is the midpoint of CB, then $SR=CR$, and so $\angle RCS\cong\angle RSC$ . Also, $\angle RSP=180-\angle RSC=180-\angle RCS$ .

Due to the midpoint theorem, PQRC is a parallelogram, which implies that $\angle RCS\cong\angle PQR$ .

In trapezoid PQRS, opposite angles $\angle RSP$ and $\angle PQR$ add up to 180. Therefore PQRS is a cyclic quadrilateral.

Now $\angle PSQ\cong\angle PRQ$ (angle subtended by arc PQ). $SP\parallel RQ$ , then $\angle PRQ\cong\angle SPR$. Then $\angle PSQ\cong\angle SPR$ . Therefore, $TS=TP$ .

Clearly, a cyclic trapezoid would be formed also by connecting the midpoint of each side and the perpendicular foot on another side.

Theorem: The quadrilateral formed by the midpoint of each side and a perpendicular foot of a triangle is a cyclic trapezoid.

# Triangle Altitudes Inequality

In $\Delta XYZ$ , the lengths of two altitudes are $a$ and $b$ , where $a>b$ . Show that the third side altitude is of range $\left(\frac{ab}{a+b},\frac{ab}{a-b}\right)$ .

Solution:
Let the sides of $\Delta XYZ$ be $s_1, s_2, s_3$ , and the corresponding altitudes be $a,b,c$, respectively.

Assuming $a>b$ implies that $s_2 > s_1$.

We then have $as_1=bs_2=cs_3 \Rightarrow c=\frac{as_1}{s_3}=\frac{bs_2}{s_3}$ .

By triangle inequality, $s_2-s_1 , and thus $\frac{1}{s_2+s_1}<\frac{1}{s_3}<\frac{1}{s_2-s_1}$ .

Writing $\frac{1}{s_2+s_1}<\frac{1}{s_3}$ in terms of the altitudes, we obtain

$\frac{bs_2}{s_2+s_1}<\frac{bs_2}{s_3} \Rightarrow \frac{b}{1+\frac{s_1}{s_2}} .

In a similar manner, it can be shown that $c< \frac{ab}{a-b}$ from $\frac{1}{s_3}<\frac{1}{s_2-s_1}$ .

Therefore, $\frac{ab}{a+b}  (Triangle Altitudes Inequality).

# Amazing Yanina

Wow.. jaw dropping haute couture. This is so far the most amazing haute couture I’ve seen this year!
I love the sophistication, elegance and ‘femininity’ with mix of sweetness and sexiness

# UAlberta Style on Campus #1

I had a chance to talk to this gorgeous fashionista in campus. She’s the first one to be featured this winter term 2015 on UAlberta Style on Campus.

“Dress for Success”

“I am in my last semester of Sociology. A lot of my friends say I overdress for school and look like a business woman, but I have been at this school for 6 years. The only thing that makes me feel better about my life sentence at the University is dressing up sometimes.”

# Office Guy

This is the perfect look I want for a guy. Guy with nice blazer, classic leather bag and hot short beard ♥

# Galaxy Pants

I went shopping 2 weeks ago. I saw these unusual designs on pants in some stores in mall. Space is really amazing, and beautiful at the same time! What can you say about the galaxy pants? 😉