Quadrilateral

[Guest]

Quadrilateral ABCD, [tex]\angle[/tex] [tex]ABD=\angle ACD=90^\circ, AD=2[/tex]. The distance between the incenters(the center of the inscribed circle in a triangle) of [tex]\Delta ABD[/tex] and [tex]\Delta ACD[/tex] is [tex]\sqrt{2}[/tex] . Find BC

[Guest]

By observation
2 of the angles are 90.
These are opposite angles.
AD = 2 is the diagonal common to both triangles
Both triangles are same size
BC is the other diagonal so is 2 also.

[Guest]

Quadrilateral ABCD, [tex]\angle[/tex] [tex]ABD=\angle ACD=90^\circ, AD=2[/tex]. The distance between the incenters(the center of the inscribed circle in a triangle) of [tex]\Delta ABD[/tex] and [tex]\Delta ACD[/tex] is [tex]\sqrt{2}[/tex] . Find BC

Tell me your are not serious with this solution please ? Did you even see what i wrote...

[Guest]

Yes, maybe I did take a simplistic viewpoint......
But, its still 2 triangles together forming a quad with 2 x 90 degree angles and a diagonal AD = 2 which is common to both triangles. .... And the inscribed circles may not be the same size.....Have to find the length of the other diagonal BC........

[Guest]

Neither BC nor AD are diagonals, it is a inscribed quad with A,B,C,D in clockwised order. Also the distance between the incenters is not given just for fun, it more complicated than that trust me.I`ll be happy if someone can offer a solution

[Guest]

|BC| = sqrt(3)

If I have a bit of time today I'll try to outline the proof in more detail, but essentially I just set A to be the origin, D to be the point at (2,0) and described the coordinates of the incenter of triangle ABD in terms of angle BAD. Repeat the process for triangle ACD using angle CAD, use pythagoras to determine the distances between the incenters which results in a relationship between the angles BAD and CAD. Write the distance BC in terms of BAD and CAD and use the previously discovered relationship to simplify the expression to sqrt(3).

Hope this helped,

R. Baber.

[Guest]

Yes, OK, you have got 2 posts since my last reply....

I was originally trying to visualise the problem as a diagram. You gave triangles ABD and ACD and angle ABD and ACD to be 90. So assumed these fitted together to form a quad. Either as a square, rectangle or a kite with inscribed circles in each triangle.
From waht you say now BC and AD are not diagonals so must be sides of the quad. So ABD and ACD are 90 and must be subtended from AD, overlapping to form clockwise ABCD formation. In my imagination this still gives a cyclic quad. ABCD in a semicircle of diameter AD = 2 and angles of 90 subtended to form triangles ABD and ACD. The inscribed circles are within each triangle and the distance between the centres is Sqrt(2) and we have to find the distance BC which is a chord drawn between the two 90 angles at the circumference. BC is a side of the cyclic quadrilateral.

If my assumptions are correct......I will have a go at providing geometrical solution.....

[leesajohnson]

BC is also diagonal of the quadrilateral so BC=2