[ASK] Find the ratio of the area of triangle BCH and triangle EHD

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Discussion Overview

The discussion revolves around finding the ratio of the areas of triangles BCH and EHD within a geometric configuration involving a parallelogram ABCD and a circle. The context includes geometric properties, similarity of triangles, and area calculations.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that triangles BCH and EHD are similar, citing specific angle measures and proposing an area ratio based on height and base lengths.
  • Another participant asserts that the area ratio can be calculated as (area of BCH)/(area of EHD) = $\dfrac{r^2}{(r\sqrt{2})^2} = \dfrac{1}{2}$, implying a specific relationship between the areas.
  • Several participants express confusion regarding the derivation of the areas of triangles BCH and EHD, seeking clarification on the calculations involved.
  • One participant provides a detailed breakdown of the area calculations, noting the bases and heights of the triangles and suggesting a simplification of the area ratio expression.
  • A final post presents a simplified ratio of 1:2 without elaboration on the derivation process.

Areas of Agreement / Disagreement

Participants express differing views on the derivation of the area ratios, with some proposing specific calculations while others seek clarification. The discussion remains unresolved regarding the exact method to simplify the area ratio.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the lengths and heights involved in the area calculations, as well as the dependence on the geometric properties of the parallelogram and circle.

Monoxdifly
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A parallelogram ABCD has angle A = angle C = 45°. Circle K with the center C intercept the parallelogram through B and D. AD is extended so that it intercepts the circle at E and BE intercepts CD at H. The ratio of the area of triangle BCH and triangle EHD is ...

Here I got that triangle BCH and triangle EHD is similar with angle BCH = angle HDE = 45°, angle CHB = angle DHE = 112.5°, and angle CBH = angle DEH = 22.5°. The area of triangle BCH is ½ × CH × h, where h is the parallelogram's height. The area of triangle EHD is ½ × (r - CH) × r. I stuck at the ratio is (CH × h) : ((r - CH) × r). How do I simplify that? Problem is, I don't know CD's length, so I can't approximate the ratio between CH and DH.
 
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A parallelogram ABCD has angle A = angle C = 45°. Circle K with the center C intercept the parallelogram through B and D. AD is extended so that it intercepts the circle at E and BE intercepts CD at H. The ratio of the area of triangle BCH and triangle EHD is ...

Note parallelogram ABCD is a rhombus with side length $r$ ...

(area of BCH)/(area of EHD) = $\dfrac{r^2}{(r\sqrt{2})^2} = \dfrac{1}{2}$
 
How did you get area of BCH and area of EHD?
 
Monoxdifly said:
How did you get area of BCH and area of EHD?

I didn't get the areas ... I determined the ratio of their respective areas.

From the diagram, let triangle BCH have base r with a perpendicular height h. Since triangle EHD is similar, its corresponding base is r√2 with corresponding height h√2.

$\dfrac{\text{area of BCH}}{\text{area of EHD}} = \dfrac{\frac{1}{2}r \cdot h}{\frac{1}{2} r\sqrt{2} \cdot h\sqrt{2}}$

now ... simplify the right side of the above equation.
 
1 : 2
 

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