Solving a Geometry Problem: Inscribed Half Circle and Triangle Relationships

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Homework Help Overview

The discussion revolves around a geometry problem involving a triangle and an inscribed half circle. The problem requires proving a relationship between the distance ED and the radius of the circle, given specific dimensions of the triangle.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the radius and the distance ED, questioning whether multiple methods exist to prove the relationship. Some express uncertainty about the need to calculate the radius, while others suggest that ED may equal half the diameter KL.

Discussion Status

The discussion is active, with participants offering various interpretations and approaches to the proof. Some have suggested that the relationship can be established without direct computation of the radius, while others are exploring more complex proofs. There is no explicit consensus on the best method to prove the relationship.

Contextual Notes

Participants note the absence of certain information, such as the center of the circle and the specific radii, which may affect the proof. The problem constraints emphasize the need for a proof rather than a calculation.

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http://members.lycos.co.uk/anzas/geo/problem.gif

in triangle ABC a half circle is inscribed.
the circle's diameter KL, is parallel to side AB
CD is the height of the triangle ABC
CD = 14 cm
AB = 10 cm
prove that ED equals to the circles radii
 
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Is there not only one radius ?
Besides, if you don't have to calculate the radius, this is pretty straightforward.
 
in the drawing there are no radii's only the diameter KL
also the center of the circle is not given.
im pretty sure that what's needed to be proven is that ED = KL/2
i just don't know how...
 
ED equal the circle radius because it is just the distance from the diameter to the parallel tangent to the circle. I don't need to calculate the radius here. Maybe you tricked me to get the answer...
 
no i didnt trick you that might be one way to prove it but there should be a proof a bit more complex...
 
Doubtlessly. Do you need to actually compute the radius ?
 
no as the question states i need to prove that ED equals to the circles radii
(in a diffrent way from the one you described)
 
Anzas said:
no as the question states i need to prove that ED equals to the circles radii
(in a diffrent way from the one you described)

Ok, humanino was right though...

since CD is the height of the triangle then CD and AB are perpendicular

which means that angle EDB = 90degrees

As well We know that a line tangent to a circle is perpendicular to the radius

In the drawing FG is perpendicular to AB so FGD = 90 degrees

as well FG and ED is perpendicular to KL since KL is parallel to AB

DEL and GFK = 90 degrees

so this proves that DEFG forms a rectangle with ED and FG being opposite sides thus being equal. FG equals the radius of the half circle, thus ED = radius of circle
 

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