Question

ABC is an equilateral triangle of side 3 units. The points P, Q lie on BC, CA re￾spectively and are such that AQ = CP = 2units. If the point R lies on AB produced

so that BR = 1unit, prove that P, Q, R are collinear

Answer 1

$AR = AB+BR = 3+1 =4$

$\text{by the cosine theorem:}$

$QR^2 = AR^2+AQ^2-2AR*AQ*\cos{\angle{A}}$

$QR^2 = 4^2+2^2-2*4*2*\cos{60^0}=12$

$QR =\sqrt{12}=2\sqrt{3}$

$\text{by the cosine theorem:}$

$QP^2 = CP^2+CQ^2-2CP*CQ*\cos{\angle{C}}$

$CQ=AC-AQ=3-2=1$

$QP^2 = 2^2+1^2-2*2*1*\cos{60^0}=3$

$QP=\sqrt {3}$

$\text{by the cosine theorem:}$

$PR^2 = BP^2+BR^2-2BP*BR*\cos{\angle{(RBP)}}$

$\angle{(RBP)}= 180^0-\angle{B}= 120^0$

$BP = BC-CP = 3-2 =1$

$PR^2=1^2+1^2-2*1*1*\cos120^0=3$

$PR= \sqrt{3}$

$QR = PR+PQ$

$\text{hence P lies strictly between R,Q}$

$\text{proven}$

Answer:$\text{proven}$