In and are the midpoints of and respectively. Lines and intersect at . If , prove that .
Connect points Q and R. Then by midpoint theorem, and .
Let M be the intersection of SB and QR.
Since SB is the height of , then MB is the height of (i.e. ). It follows that M is the midpoint of SB and by SAS postulate. Hence, . As R is the midpoint of CB, then , and so . Also, .
Due to the midpoint theorem, PQRC is a parallelogram, which implies that .
In trapezoid PQRS, opposite angles and add up to 180. Therefore PQRS is a cyclic quadrilateral.
Now (angle subtended by arc PQ). , then . Then . Therefore, .
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.
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
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.”
This is the perfect look I want for a guy. Guy with nice blazer, classic leather bag and hot short beard ♥
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? 😉
Here are some of my favorite styles by a young stylist in Manila: Camille Co
“Androgyny Goes To School”
More styles on http://www.chictopia.com/camilleco