# Prove that $BP,CQ,AD$ concur

Triangle $$ABC$$ has an incircle $$(I)$$ which contacts $$BC,CA,AB$$ at $$D,E,F$$. On line $$EF$$ we get two points $$M$$ and $$N$$ such that $$CM//BN//AD$$. $$DM$$ and $$DN$$ cut $$(I)$$ at $$P,Q$$.

a, Prove that $$BP,CQ,AD$$ concur.

b, Let $$J$$ be point which $$BP,CQ,AD$$ concur. $$X$$ is midpoint of $$PQ$$. Show that $$JX$$ intersects $$MN$$ at the midpoint $$G$$ of $$MN$$.

I don't know which lemmas we use(maybe Ceva theorem, Thales theorem because there are three paralel lines). Show please and anyone can tell me some geometry book for studying? Thank. • You mean $JX$ goes through $MN$ at $G$ and $G$ is midpoint of $MN$? – Word Shallow Jan 1 at 4:27
• Yes, of course. help me show this pls. – Quỳnh Vũ Thị Jan 1 at 5:42

## 1 Answer

a) Let $$S$$ be the intersection of $$EF$$ and $$BC$$,the segment $$AD$$ and incircle $$(I)$$ be $$T$$; $$J$$ be the intersection of $$SP$$ and $$AD$$; $$BQ$$ intersects $$CP$$ at $$V$$ .Then we have $$ST$$ is tangent of incircle $$(I)$$

So the polar of $$S$$ is the line $$AD$$.

We have: $$V(SJ,QP)=-1$$ and $$V(SD,BC)=-1$$

And $$VS\equiv VS,VB\equiv VQ,VC\equiv VP\Rightarrow VD\equiv VJ$$

So $$V\in AD$$. In $$\Delta VBC$$: $$P;Q$$ are respectively in the $$CV$$ and $$BV$$

$$PQ$$ intersects $$BC$$ at $$S$$ and $$(SD,BC)=-1$$ so we have $$MV;BQ;CQ$$ concur.

Or $$BP;CQ;AD$$ concur $$(Q.E.D)$$

• What about exercise b? Thank you. – Quỳnh Vũ Thị Jan 2 at 3:50