Need help with a triangle similarity problem: Triangle CBD is inscribed in a circle. From point D, a tangent line is drawn...

[rocketFrog]

Hi everyone,

Problem statement:
Triangle CBD is inscribed in a circle. From point D, a tangent line is drawn, which intersects the extension of side CB at point E. Two normals from the points B and C are drawn onto the tangent line, the shorter of which is 6cm in length. Calculate the area of the trapezoid formed by the two normals, side CB and a part of the tangent if CB = 5cm, DE = 5[imath]\sqrt{6}[/imath] cm and BF = 6cm.


The formula for the area of a trapezoid is [imath]A = \frac{(a+c)h}{2}[/imath] where a and c are the bases and h is a height. Now, the trapezoid in question is a right trapezoid, if we flip the image for 180º, we have:

So we already have the longer side CB, and shorter base FB, to calculate the area we need the lengths of longer base GC and height GF.
Triangles GEC and FEB since they both have a 90º angle and they share the angle at point E.

This is pretty much where I'm stuck. I've tried looking at triangles FEB and FDB but was not able to prove similarity between the two.

Any help would be appreciated!!

 

[Dr.Peterson]

Hi everyone,

Problem statement:
Triangle CBD is inscribed in a circle. From point D, a tangent line is drawn, which intersects the extension of side CB at point E. Two normals from the points B and C are drawn onto the tangent line, the shorter of which is 6cm in length. Calculate the area of the trapezoid formed by the two normals, side CB and a part of the tangent if CB = 5cm, DE = 5[imath]\sqrt{6}[/imath] cm and BF = 6cm.

View attachment 36929
The formula for the area of a trapezoid is [imath]A = \frac{(a+c)h}{2}[/imath] where a and c are the bases and h is a height. Now, the trapezoid in question is a right trapezoid, if we flip the image for 180º, we have:
View attachment 36930
So we already have the longer side CB, and shorter base FB, to calculate the area we need the lengths of longer base GC and height GF.
Triangles GEC and FEB since they both have a 90º angle and they share the angle at point E.

This is pretty much where I'm stuck. I've tried looking at triangles FEB and FDB but was not able to prove similarity between the two.

Any help would be appreciated!!

You might look for a pair of similar triangles, or for a theorem relating EB, EC, and ED (whose proof involves those same similar triangles), depending on what theorems you know. This can give you EB; I did some further work with the same similar triangles and their areas, but there may well be other ways.