Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 65Right Triangle, Midpoints. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
Let B be (0,0) BE the x-axis & BA the y-axis. Further, let BE=p and BD=q. This gives, E:(p,0), C:(p+e,0),D:(0,q), A:(0,d+q). Thus N:((p+e)/2,(q+d)/2)and M;(p/2,q/2). By the distance formula, MN^2 = x^2 = ((p+e)/2 -p/2)^2+((q+d)/2 -q/2)^2 = e^2/4 + d&2/4 0r x^2 =(1/4)*(d^2+e^2)or x=(1/2)*(d^2+e^2)^(1/2)Ajit: firstname.lastname@example.org
I like the geometric solution Joe so I got this solution : Let K and L be the orthogonal projections of A and C on (DE), take S the midpoint of [KL] such that SM=p now KBEA and DBLC are concyclic so the two triangles AKD and CLE are similar then LC*KA=LE*KD-------(1)now p=(KM-LM)/2 and the median NS=(CL+AK)/2Now x^2=(NS)^2+P^2 =(KM^2-2KM*LM+LM^2+CL^2+2CL*AK+AK^2)/4 Let DM=ME=yso x^2=((KD+y)^2-2(KD+y)(LE+y)+(LE+y)^2+CL^2+2CL*AK+AK^2)/4 = (CL^2+AK^2+KD^2+2yKD+y^2+LE^2+2yLE+y^2-2KD*LE-2yKD-2yLE-2y^2+2CL*AK)/4 =(CL^2+AK^2+KD^2+LE^2-2KD*LE+CL*AK)/4 =(e^2+d^2)/4 (from (1) but this solution is long Is there shorter Antonio????? why there is no post for problem 358 the solution is already prepared and I have suggest for you make an area in your site to our problems if we have because I have many problems didn't mention in this with these problems thx a lot
Extend AM to P such that AM = MP so ADPE is a parellelogram and Tr. PEC is a right Tr. with EP = dNow extend MN to Q such that MN = NQ = x. So AMCQ is a //ogram with QS = AM = MP. Hence MPCQ is a //ogram and PC = MQ = 2x.Now apply Pythagoras to right Tr. CEP and d^2 + e^2 = 4x^2 and so the result follows. Sumith PeirisMoratuwaSri Lanka
Variation of this problemIf B = 60 and d = e then d = 2x