2 thoughts on “2017 Paper a Day 51: Sydney Grammar Ext 2 2015”
Hi,
“the join of the midpoints of two sides of a triangle is parallel to and equals half the third side”
Can we just quote this theorem in a question?
I’ve seen it appear twice now in Terry Lee’s hsc solutions. I’m currently looking at Qu 5c iii) for 2011 HSCExt2 and i can’t see any other way to do this qu.
Thanks
It is not a Course Theorem, so NO you cannot just quote it , if you want to use it , you should prove it (using similar triangles)
I think the neatest proof is;
slope AQ X slope A’Q = y/(x-a) X y/(x+a) = y^2/(x^2-a^2)
If Q lies on the circle then slope AQ X slope A’Q = -1 (angle in semicircle = 90)
i.e. y^2/(x^2-a^2) = -1
y^2=a^2-x^2
x^2+y^2=a^2 which is the equation of the circle
thus Q lies on the circle
Hi,
“the join of the midpoints of two sides of a triangle is parallel to and equals half the third side”
Can we just quote this theorem in a question?
I’ve seen it appear twice now in Terry Lee’s hsc solutions. I’m currently looking at Qu 5c iii) for 2011 HSCExt2 and i can’t see any other way to do this qu.
Thanks
It is not a Course Theorem, so NO you cannot just quote it , if you want to use it , you should prove it (using similar triangles)
I think the neatest proof is;
slope AQ X slope A’Q = y/(x-a) X y/(x+a) = y^2/(x^2-a^2)
If Q lies on the circle then slope AQ X slope A’Q = -1 (angle in semicircle = 90)
i.e. y^2/(x^2-a^2) = -1
y^2=a^2-x^2
x^2+y^2=a^2 which is the equation of the circle
thus Q lies on the circle