Triangle ABC is right-angled at A, and AB < AC. A point H is on the segment AC (H is different from A and C). Denote E as the perpendicular projection of H onto BC. Prove that angle HBC = angle EAC.

You are very welcome. Turns out that there a whack load of similar triangles in the diagram and I wasted time on those. The breakthrough for me was getting the 2 right triangles in between the 2 short parallel lines, HEG and HEF similar, which in turn gave me the equal ratios that I needed.

If you have the time, you should check out the properties of circles. Specifically, the sector angle is 2x the inscribed angle subtended by the same arc. This leads to all inscribed angles, subtended by the same arc, being equal. Very powerful theorems.

It took me a while - equal angles everywhere. But the 2 constructions saved the day. They allowed me to use the "proportionality theorem" , "angle-angle-angle similarity" and finally

"side-angle-side similarity" to get the desired angles equal.

So there's apparently a bunch of spammers running wild on the website. For now don't trust anything with a link in it that has nothing to do with the post.

@Setsuna Kujo I have notified bprp on several occasions through "Let's Chat". He responded once - claims he can't stop it . All he needs to do is monitor this board every couple of days and just DELETE the spam. Can't take more then 5 min. I don't understand why he doesn't do it. I suggest you all use the chat and tell him. Let's see if we can get this under some kind of control. It's drowning this board.

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I don't know if this could help you or not, but I want to clarify that the proportionality theorem doesn't just apply to scalene triangles, it also applies to other triangles as well. My mistake.

Roger that. Know the theorem well. Similar Triangles are one of the basic and fundamental building blocks of all of mathematics. I will give this a closer look.

Yes, but we are supposed to solve it without any knowledge about cyclic quads or anything about circles at all. We were supposed to solve it using knowledge about the proportionality theorem on triangles and the three cases of similar triangles. Can you try to do another solution for that?

You are very welcome. Turns out that there a whack load of similar triangles in the diagram and I wasted time on those. The breakthrough for me was getting the 2 right triangles in between the 2 short parallel lines, HEG and HEF similar, which in turn gave me the equal ratios that I needed.

If you have the time, you should check out the properties of circles. Specifically, the sector angle is 2x the inscribed angle subtended by the same arc. This leads to all inscribed angles, subtended by the same arc, being equal. Very powerful theorems.

It took me a while - equal angles everywhere. But the 2 constructions saved the day. They allowed me to use the "proportionality theorem" , "angle-angle-angle similarity" and finally

"side-angle-side similarity" to get the desired angles equal.

Have you got any idea?

Capstone Writing Service??? BPRP - can you deep 6 these commercials which clutter up the discussion - please.

I am so grateful for your help. You are the

best capstone project writing service. I wish you good luck, Godspeed and farewell. Please keep it good posting!Is this the correct diagram?

Roger that. Know the theorem well. Similar Triangles are one of the basic and fundamental building blocks of all of mathematics. I will give this a closer look.

What exactly is the "Proportionality theorem" ?

Quad ABEH is cyclic (opposite <'s add to 180 deg)

Therefore <HBE = <EAH (Inscribed Angle Theorem)

Since both BEC and AHC are straight, then <HBC = <EAC