# Answer in Geometry for Tahiru Rabiu #210714

An equilateral triangle ABC has coordinates O(0, 0), B(a, 0) and C(c, d).

a Find c and d in terms of a by using the fact that OB = BC = CO.

b Find the equation of the lines OB, BC and CO.

In any triangle PQR prove that PQ2 + PR2 = 2(PS2 + SR2)

Where S is the midpoint of QR.

Prove that the midpoints of a parallelogram bisect each other using coordinate geometry.

Solution

1.

a)

“OB = BC = CO”

“OB^2 = BC^2 = CO^2”

“a^2=(a-c)^2+(-d)^2=c^2+d^2”

“a^2-2ac+c^2=c^2”

“c=dfrac{a}{2}, anot=0”

“a^2=dfrac{a^2}{4}+d^2”

“|d|=dfrac{sqrt{3}}{2}|a|”

Let “a>0, d>0.” Then

“c=dfrac{a}{2}, d=dfrac{sqrt{3}}{2}a”

b)

“OB: y=0”

“BC: y=-sqrt{3}x+sqrt{3}a”

“CO:y=sqrt{3}x”

2.

Let “P(0,0), Q(2a, 2b), R(2c, 2d).” Then “S(a+c, b+d).”

“PQ^2=(2a)^2+(2b)^2”

“PR^2=(2c)^2+(2d)^2”

“PS^2=(a+c)^2+(b+d)^2”

“SR^2=(2c-(a+c))^2+(2d-(b+d))^2”

Substitute

“2(PS^2+SR^2)=2PS^2+2SR^2”

“=2((a+c)^2+(b+d)^2)+2((c-a)^2+(d-b)^2)”

“=2(a^2+2ac+c^2+b^2+2bd+d^2)”

“+2(c^2-2ac+a^2+d^2-2bd+d^2)”

“=4a^2+4b^2+4c^2+4d^2”

“=(2a)^2+(2b)^2+(2c)^2+(2d)^2”

“=PQ^2+PR^2”

Therefore

“PQ^2+PR^2=2(PS^2+SR^2)”

3. Let “ABCD” be a parallelogram

“A(0, 0), B(b, h), C(b+d, h), D(d, 0)”

“BCparallel AD, AD=BC”

Then

“AC: y=dfrac{h}{b+d}x”

“BD: y=dfrac{h}{b-d}x-dfrac{hd}{b-d}”

Intersection

“dfrac{h}{b+d}x=dfrac{h}{b-d}x-dfrac{hd}{b-d}”

“x(bh-dh)=x(bh+dh)-bdh-d^2h”

“2dhx=bdh+d^2h”

“x=dfrac{b+d}{2}”

“y=dfrac{h(b+d)}{2(b+d)}=dfrac{h}{2}”

Point “E(dfrac{b+d}{2}, dfrac{hd}{2})” is the midpoint of the diagonal “AC.”

Point “E(dfrac{b+d}{2}, dfrac{hd}{2})” is the midpoint of the diagonal “BD.”

Therefore the diagonal of a parallelogram bisect each other.