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.
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?
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.
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.
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
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?
What exactly is the "Proportionality theorem" ?
Let's say you have a normal, scalene triangle ABC. You then pick a point D on AB and E on AC such that DE || BC. Then, we can say that AD/AB = AE/AC.
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.
Is this the correct diagram?
Yes, exactly.
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.
@Minh Flip Bottle It applies to all triangles.
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