# circles, power of point, cross ratios

Let $w$ be a circle, and let $P$ be a point outside $w$. Let $X, Y$ be the tangents from $P$ to $w$. A line from $P$ intersects $w$ in two points $B, D$. Let $C$ be the intesection of $\overline{XY}$ and $\overline{BC}$. Prove that $CD\cdot BP=BC \cdot DP$.

I've tried using power of the point repeatedly but just get a mess. Cross ratios can be computed for four collinear points (and the relation we are asked to prove is that the cross ratio (ignoring signs) is $1$) but I don't know how to use that.

• $BC\cdot CD=CX\cdot CY$ – Lucian Dec 14 '15 at 6:44

Denote $PX=PY=t$, $CD=a, BC=b, PD=c, CX=x, CY=y$.

By the power of point $C$ we have $$ab=xy.$$

By the power of point $P$ we have

$$t^2=c(a+b+c).$$

Finally, apply Stewart's theorem in triangle $PXY$ for the cevian $PC$. We obtain:

$PC^2\cdot XY+CX\cdot CY \cdot XY=PX^2\cdot CY+PY^2\cdot CX$ from which

$(c+a)^2\cdot(x+y)+xy(x+y)=t^2(x+y)$ and after dividing by $x+y$ it follows that

$$t^2=(c+a)^2+xy.$$

Combining the three equations above we obtain $a(a+b+c)=bc$ which is the desired equality.

• I have not studied Stewart's theorem.Please tell me Stewart's theorem is part of which standard. – mathlover Dec 14 '15 at 14:01
• @Ayush Jaiswal Stewart's theorem can be easily proved by using the law of cosines twice. See en.wikipedia.org/wiki/Stewart%27s_theorem. – user84909 Dec 14 '15 at 14:04