Geometry – Length Chasing 1

Geometry Level 4

Triangle $ABC$ is right angled at $C$ . The lines $AP,BQ$ and $CS$ intersect at point $R$ . $AB$ extended and $QP$ extended meet at $T$ . If $AB=10, AC=8, CQ=2$ and $CP=2$ , find $TS$ .

Note: The figure is not true to scale.

The answer is 24.

1 solution

A Former Brilliant Member
Oct 22, 2017

By Ceva’s Theorem, we have

$\dfrac{AQ}{QC} \times \dfrac{CP}{PB} \times \dfrac{BS}{SA}=1$

$\dfrac{6}{2} \times \dfrac{2}{4} \times \dfrac{BS}{10-BS}=1$

$\dfrac{3BS}{2(10-BS)}=1$

$BS=4$

By Menalaus’ Theorem, we have

$\dfrac{TA}{TB} \times \dfrac{PB}{CP} \times \dfrac{CQ}{AQ}=1$

$\dfrac{TB+10}{TB} \times \dfrac{4}{2} \times \dfrac{2}{6} = 1$

$TB=20$

It follows that,

$TS=TA-6=20+10-6=$ $\boxed{24}$

Geometry – Length Chasing 1

The answer is 24.

1 solution

0 pending reports