In triangle ABC medians of triangle BE and AD are perpendicular to each other. Find the length of ¯AB, if ¯BC=6 and ¯AC=8
Problem
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Tags: Pythagorean Theorem, geometry, geometry proposed
13.03.2011 22:48
Let the intersection of AD,BE be O. Let OE=e,OD=D. Then OA=2d,OB=2e. By the Pythagorean Theorem, e2+4d2=AE2=16,d2+4e2=DB2=9, adding which yields d2+e2=5. Thus, AB=√4d2+4e2=√4(d2+e2)=√4(5)=2√5.
30.12.2019 18:21
BigSams wrote: Let the intersection of AD,BE be O. Let OE=e,OD=D. Then OA=2d,OB=2e. By the Pythagorean Theorem, e2+4d2=AE2=16,d2+4e2=DB2=9, adding which yields d2+e2=5. Thus, AB=√4d2+4e2=√4(d2+e2)=√4(5)=2√5. Could you tell me why, d2+e2=5
22.12.2020 15:24
@above Because the point where two medians intersect has to be the centroid and the centroid divides a median in the ratio 2:1
22.12.2020 17:56
I am going to BARY this problem. Use Barycentrics on △ABC and set A=(1,0,0),B=(0,1,0),C=(0,0,1),D=(0,12,12),E=(12,0,12). We have →AD=(−1,12,12) and →BE=(12,−1,12). By EFFT, →AD⊥→BE⟺62(14−12)+82(−12+14)+c2(1+14)=0. Solving for c, our answer is 2√5. ◼
30.04.2023 15:09
Let O be the intersection of AD,BE Let OE=a and OD=b ⇒OB=2a,OA=2b Using pythagorean theorem, we get: a2+(2b)2=AE2⇒a2+4b2=42=16 b2+(2a)2=BD2⇒b2+4a2=32=9 Adding them together, we get: 5a2+5b2=25⇒a2+b2=5 Finally, we get: AB2=(2a)2+(2b)2=4a2+4b2=4(a2+b2)=4⋅5=20⇒AB=√20=2√5 ◼