Prove that $AK$ passes through passes the midpoint of $DE$.

Consider $$AD$$, $$BE$$ and $$CF$$ being altitudes of $$\triangle ABC$$ where $$D \in BC$$, $$E \in CA$$ and $$F \in (A, B, C)$$. $$FD$$ extended intersects $$(A, B, C)$$ at $$K$$. Prove that $$AK$$ passes through the midpoint of $$DE$$.

Here's what I've done.

Let the intersections of the tangent of $$(A, B, C)$$ at $$C$$ and $$AK$$, $$AD$$ respectively $$G$$ and $$H$$.

We have that $$\widehat{GCK} = \widehat{GAC} \implies \triangle GCK \sim \triangle GAC \implies \dfrac{GC}{GA} = \dfrac{GK}{GC} \implies GK \cdot GA = GC^2$$

Now I just need to prove that $$GD^2 = GK \cdot GA$$ so that $$GC = GD \implies \widehat{GCD} = \widehat{GDC}$$

$$\implies \widehat{GHD} = \widehat{GDH} \implies GH = GD \implies GC = GH$$.

Then I will let the intersection of $$AK$$ and $$DE$$ be $$I$$.

We have that $$AEDB$$ is a cyclic quadrilateral because $$\widehat{ADB} = \widehat{AEB} = 90^\circ$$.

$$\implies \widehat{BAE} + \widehat{BDE} = 180^\circ \implies \widehat{BAC} = \widehat{CDE}$$.

And $$\widehat{BAC} = \widehat{BCH}$$ because $$CH$$ is a tangent of $$(A, B, C)$$ at $$C$$.

$$\implies \widehat{CDE} = \widehat{BCH} \implies DE \parallel CH$$.

Using the intercept theorem for $$DI \parallel HG$$ and $$EI \parallel CG$$.

We have that $$\dfrac{DI}{HG} = \dfrac{AI}{AG}$$ and $$\dfrac{EI}{CG} = \dfrac{AI}{AG}$$ $$\implies \dfrac{DI}{HG} = \dfrac{EI}{CG}$$.

But $$HG = CG \implies DI = EI$$.

• What do you want?
– Aqua
Apr 5 '19 at 15:56
• Nothing. The question describes my attempt. The answer, well..., answers the question. Apr 5 '19 at 16:01

Answer brought to you by my teachers. The story is:

This problem is my homework from someone who shall not be named that teaches me in extra classes - voluntary classes you would go to when you are done studying at school, whether it is to revise things you learn at school or to acquire new knowledge that you don't at school.

I was studying this afternoon at school with another who shall not be named. I was done figuring out geometry problems at the chalkboard so I asked this problem to my teacher at school in the first period. And she gave me back the solution the next period and said

"Don't ever give me this kind of problem again. It is too difficult for me."

I knew what happened right then. My two teachers are co-workers, so one asked the other for clues. I have to say that both teachers are talented although this problem and solution could be pulled out of a reference without me knowing.

I know the story is redundant and against the rules of the site but I just want to share this story. I have put personal values above others, please forgive me.

Now back to the problem.

Let $$M$$ be the midpoint of $$DE$$ and $$AB \cap FH = G$$. $$H$$ is the orthocenter of $$\triangle ABC$$.

We need to prove that $$\widehat{EAM} = \widehat{CAK}$$.

First, we have that $$\widehat{FBA} = \widehat{FCA}$$ and $$\widehat{HBA} = \widehat{HCA} = (90^\circ - \widehat{CAB})$$ $$\implies \widehat{FBA} = \widehat{HBA}$$.

That means $$F$$ and $$H$$ reflect one another in line $$AB$$ $$\implies HF = 2HG$$.

$$M$$ is the midpoint of $$DE$$ $$\implies ED = 2EM$$ $$\implies \dfrac{HF}{ED} = \dfrac{HG}{EM} \implies EM \cdot HF = ED \cdot HG$$.

Because $$\widehat{ADB} = \widehat{AEB} = 90^\circ$$ and $$\widehat{HDB} = \widehat{HGB} = 90^\circ$$, $$AEDB$$ and $$HDBG$$ are cyclic quadrilaterals.

It's easy to prove that $$\widehat{DGH} = \widehat{DAE} \ (= \widehat{DBE})$$ and $$\widehat{GHD} = \widehat{AED} \ (= 180^\circ - \widehat{DBA})$$.

$$\implies \triangle HDG \sim \triangle EDA \implies \dfrac{HD}{ED} = \dfrac{HG}{EA} \implies EA \cdot HD = ED \cdot HG$$

$$\implies EA \cdot HD = EM \cdot HF \implies \dfrac{HD}{EM} =\dfrac{HF}{EA} \implies \triangle HDF \sim \triangle EMA \implies \widehat{DFH} = \widehat{MAE}$$.

But $$\widehat{DFH} = \widehat{KAC}$$ $$\implies \widehat{KAC} = \widehat{MAE}$$.

That means $$AK$$ passes through point $$M$$, the midpoint of $$DE$$.