Prove that the following lines so described are perpendicular.

Question

If O be any point on the circumcircle of the triangle ABC, and OL be drawn parallel to BC to meet the circumcircle in L. OD and OF are drawn perpendicular to BC and AB respectively, then show that LA will be perpendicular to DF.

The above question is from my self-preparatory material for an exam(JEE). I tried a lot thinking on this but I don't know from which theorem or property of triangles and circles to start with. Any suggestion on how to tackle such problems will be appreciated.

Answer 1

In quadrilateral AFGH we have:

$\angle AHG= \frac{Arc{GCA}}2$

$\angle AFG= \frac{Arc{AOG}}2$

Summing sides we get:

$\angle AHG+\angle AFG=\frac{Arc{GCA}}2+\frac{Arc{AOG}}2=180^0$

$\angle HGF= \frac{Arc{HOF}}2$

$\angle HAF= \frac{Arc{FCB}}2$

Summing both sides we get:

$\angle HGF+\angle HAF=\frac{Arc{HOF}}2+\frac{Arc{FCB}}2=180$

That is the sum of opposite angles of quadrilateral AFGH are equal which means it is an isosceles trapezoid which in turn means AH is parallel with FG. But $\angle HAL =90^o$(can you find why?) so $\angle DEL=90^o$ and this is what we had to show.

Additional observation: If B moves to H such that condition OL||BC reserved , B, D and H will be coincident. In this case while B moves on circle, D moves on extension of OH such that HD keeps being perpendicular on BC. Since HA is perpendicular on AL, it is reasonable to accept that AH keep being perpendicular on AL in its new positions including DF. In fact DF is transform of AH when B moves from H to it's own position.This shows that DE||AH.

Answer 2

I already saw one solution above but I decided to add my solution because why not. :)

Drop a perpendicular from $O$ to segment $AC$. Then points $D, E$ and $F$ are collinear as $DEF$ is Simson's line.

Now I claim that $∠OFP = ∠FAP$.

Note that,

$∠OFP = ∠OFE = ∠OAE = ∠OAC = ∠OLC$ as $FAEO$ is a cyclic quadrilateral and $∠OAC$ and $∠OLC$ subtends minor arc $OC$.

Also note,

$∠FAP = ∠LAB = ∠LCB = ∠OLC$ as $ALBC$ is cyclic and $OL ∥ BC$.

Thus $∠OFP = ∠FAP$. As claimed.

Now notice, $∠OFA = ∠OFP + ∠PFA = 90$. But by the claim, $∠PFA + ∠FAP = 90$

$\implies$ $∠PFA = 90 - ∠FAP$. Thus $\triangle FAP$ is a right triangle. Hence, the result.