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.
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