Question

$A$ line $L$ intersects three sides $BC,CA$ and $AB$ of a triangle in $P,Q,R$ respectively, show that
$\frac{BP}{PC}.\frac{CQ}{QA}.\frac{AR}{RB}=-1$

Answer 1

The proof your question requires seems wrong.Construct perpendiculars AD, CH and BI from A, B and C to DEF.The triangles △BRG and △ABH wil be similar because both are right angled triangleson same base.Hence △BRG~△ABH⇒∣∣ABRB∣∣=∣∣AHRG∣∣ ;(i)Similarly△QIC~△ACH⇒∣∣CQAC∣∣=∣∣QIAH∣∣ ;(ii)△QIP~△RPG⇒∣∣PRPQ∣∣=∣∣RGQI∣∣ ;(iii)Multiplying three equations we get,∣∣ABRB×CQAC×PRPQ∣∣=∣∣AHRG×QIAH×RGQI∣∣⇒∣∣ABRB×CQAC×PRPQ∣∣=1Now the line DEF must intersect either two sides of △ABC.That means there is an odd number of negative contributions to the product. Hence the equation works out to be negative.⇒(AB/RB)×(CQ/AC)×(PR/PQ)=−1