blackpenredpen

math for fun 

To see this working, head to your live site.
  • Categories
  • All Posts
  • My Posts
Minh Flip Bottle
Jun 11, 2021

Geometry problem

in Math Problems

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.

18 comments
0
Ian Fowler
Jun 11, 2021  ·  Edited: Jun 11, 2021

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

0
Minh Flip Bottle
Jun 12, 2021

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?

0
Ian Fowler
Jun 12, 2021

What exactly is the "Proportionality theorem" ?

0
Minh Flip Bottle
Jun 13, 2021

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.

0
Ian Fowler
Jun 13, 2021

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.

Ian Fowler
Jun 13, 2021

Is this the correct diagram?




0
Minh Flip Bottle
Jun 14, 2021

Yes, exactly.

0
Minh Flip Bottle
Jun 16, 2021

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.


0
Ian Fowler
Jun 16, 2021

@Minh Flip Bottle It applies to all triangles.