Question

Geometry Problem

Answer 1

I just want to know, how will we prove it when AC = AD.

Answer 2

This was the solution for AC not = AD,


R=DA, r = CA, 2a = AB(where O is the center)
Join B and D ->Triangle BDA is right triangle
Join B and C ->Triangle BCA is right triangle

Draw a radius DF (DF is perenpedicular to MN)
Draw a radius CE (CE is perenpedicular to MN)
Draw an tangent to circle O threw A in that way that it cut DF at L and CE at M thus MEFL is rectangle.

Triangle DAL is similiar to BDA and so MAC to BCA (because they has 90 angle of MEFL and becasue of the tangent : R/2a = DL/R (triangles DAL and BAD) - >DL = R^2/2a THUS LF = R - R^2/2a
AC/BA = MC/AC -> r/2a = MC/r (triangles CAM and BCA) - >MC = r^2/2a THUS ME = r - r^2/2a

Since MEFL is rectangle we get that ME = LF:
r - r^2/2a = R - R^2/2a
after short develop we get
2ar - r^2 = 2aR - R^2
R^2 - r^2 = (R - r)(R+r) = 2a(R - r)
thus 2a = R+r - > AB = DA + AC


But, if you look at the step,

R^2 - r^2 = (R - r)(R+r) = 2a(R - r)
thus 2a = R+r - > AB = DA + AC

If, R would have been equal to r, then the the required proof would have been wrong.

Answer 3

Thus, I am looking for some proof if AC = AD.

Is the required equation to prove even true for R = r ?

Answer 4

This seems easy! :P

@mathmale @iambatman @zzr0ck3r may help you!

Answer 5

Not one geometry class in my undergrad....sorry

Answer 6

No problem. @zzr0ck3r Though, thanks for looking through the problem.