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

[Monoxdifly]

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.

 

[skeeter]

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) = $\dfrac{r^2}{(r\sqrt{2})^2} = \dfrac{1}{2}$

 

[Monoxdifly]

How did you get area of BCH and area of EHD?

 

[skeeter]

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.

 

[Monoxdifly]

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