Prove that perpendiculars from D,E,F to BC,CA,AB are concurrent?

Question

Question -

If ABC and DEF be any two triangles such that the perpendiculars from the points A,B,C to the sides EF,FD,DE respectively, are concurrent. Prove that the perpendiculars from D,E,F to BC,CA,AB respectively are also concurrent.

My attempt -

First of all I know this question has already been answered but there they used carnots theorem and some other things which I haven't learn yet...

I am not able to figure it out using ceva theorem...I draw two triangles forming STAR ...but did not get anything worth..then I draw both triangle on same plane but then things become very confusing...

I want some hint or how to draw correctly this thing... Thankyou

Answer 1

Let $AA_1,$ $BB_1$, $CC_1,$ $DD_1$, $EE_1$ and $FF_1$ be our perpendiculars.

Thus, $$\measuredangle(AA_1,AC)=\measuredangle(EE_1,FE),$$ $$\measuredangle(AA_1,AB)=\measuredangle(FF_1,FE),$$ $$\measuredangle(FF_1,FD)=\measuredangle(BB_1,AB),$$ $$\measuredangle(BB_1,BC)=\measuredangle(DD_1,FD),$$ $$\measuredangle(DD_1,DE)=\measuredangle(CC_1,BC)$$ and $$\measuredangle(CC_1,AC)=\measuredangle(EE_1,DE).$$ Thus, our statement follows from the Ceva's theorem in the trigonometric form.

Answer 2

This is not the answer to your question. This is the depicture of your problem description, which you had trouble in drawing. I hope this will help you to find an appropriate answer.