# Angle chasing to show three points are collinear.

Let $$ABC$$ be an acute triangle with circumcenter $$O$$ and let $$K$$ be such that $$KA$$ is tangent to the circumcircle of $$\triangle ABC$$ and $$\angle KCB = 90 ^{\circ}$$. Point $$D$$ lies on $$BC$$ such that $$KD || AB.$$ Show that $$DO$$ passes through $$A.$$

This is a problem from EGMO, eg.1.32.

## My approach:

I created a dummy point D' such that D'O passes through A. Now I just have to show that BA and D'A are || which will solve the problem. I think I have all the necessary theorems and I think this problem can be solved with angle chasing, but I can't really figure out how to proceed.

• thanks @YNK...... Sep 25 at 4:27

We shall try to work backwards in order to reach the desired conclusion.

Let $$\triangle ABC$$ be an acute triangle with circumcenter $$O$$ and let $$K$$ be such that $$KA$$ is tangent to the circumcircle of $$\triangle ABC$$ and $$\angle KCB=90^{\circ}.$$

Define $$L$$ to be any point on line $$KA$$ such that $$A$$ lies between $$K$$ and $$L$$. Also, extend segment $$AO$$ to point $$X$$ on $$BC$$.

Using Alternate Segment Theorem, $$\angle BAL=ACB=\alpha\implies \angle AOB=2\alpha\implies \angle OAB=\angle BAX=90-\alpha.$$ Since, $$\angle XAL=\angle XCK=90^{\circ}, AKCX$$ is cyclic and hence, $$\angle AKX=\angle ACX=\alpha\implies AB\parallel KX.$$

But there exists a unique point $$D$$ on $$BC$$ such that $$AB\parallel KD$$. Therefore, $$X\equiv D$$ and $$A, O, D$$ are collinear.

It is sufficient to show that $$\angle CAO=\angle CAD$$.

Let the tangent be labelled $$KAK'$$. Then due to the Alternate Segment Theorem, $$\angle K'AB=\angle ACB$$ And since $$AB\parallel KC$$, both of these equal $$\angle AKD$$. The tangent is perpendicular to the radius so $$\angle BAO=90^o-\angle K'AB$$.

But from the data given, $$\angle KCA=90^o-\angle ACB\implies \angle KCA=\angle BAO$$

Considering the total angle in triangles $$ABC$$ and $$AKC$$, the remaining angle in each is the same, so $$\angle OAC=\angle DKC$$

However, points $$AKC$$ and $$D$$ are concyclic since $$\angle AKD=\angle ACD$$ (due to Angles in the Same Segment are Equal).

Therefore, for the same reason, $$\angle DAC=\angle DKC$$.

Hence, finally, $$\angle CAO=\angle CAD$$ QED