geometry
A median of a triangle is a line segment from a vertex of a triangle to the midpoint of the opposite side of the triangle. The medians to the legs of a certain right triangle have lengths 13 and 19. What is the length of the hypotenuse of the triangle?
I don't know how to find the length of the hypotenuse with such little information given. Any ideas?
This is my short proof from 1963:
In the triangle ABC draw medians BE, and CF, meeting at point G. Construct a line from A through G, such that it intersects BC at point D.
We are required to prove that D bisects BC, therefore AD is a median, hence medians are concurrent at G (the centroid).
Proof:
Produce AD to a point P below triangle ABC, such that AG = GP.
Construct lines BP and PC.
Since AF = FB, and AG = GP, FG is parallel to BP. (Euclid)
Similarly, since AE = EC, and AG = GP, GE is parallel to PC
Thus BPCG is a parallelogram.
Since diagonals of a parallelogram bisect one another (Euclid), therefore BD = DC.
Thus AD is a median. QED
Corollary: GD = AD/3.
Proof:
Since AG = GP and GD = GP/2, AG = 2GD.
AD = (AG + GD) = (2GD + GD) = 3GD.
Hence GD = AD/3. QED
Refer to the figure below.
Let A be (2, -4) and B be (p, q).
L(1) : 2x-3y-2=0
L(2): 5x+3y-12=0
Testing showed that A is not on either L(1) or L(2).
Note: the next 6 lines can be skipped.
Solving the above, we have G, the centroid is at (2. 2/3).
This means A and G are on the same vertical line x = 2.
AG = (2/3) + 4 = 14/3
Let A(h) and C(h) be the feet of the medians through A onto BC and C onto AB respectively.
By the properties of centroid, GA(h) = 0.5*AG = 7/3
A(h) is then at (2, 3).
The midpoint of AB = C(h) = ([p + 2]/2, [q – 4]/2).
It lies on L(2). Therefore, 5[[p + 2]/2] + 3[[q – 4]/2] -12 = 0………(1)
B(p, q) lies on L(1). Therefore, 2(p) – 3(q) – 2 = 0……………………..(2)
Solving (1), and (2), we have B = (4,2).
The co-ordinates of C can be similarly obtained.
The required equations can be obtained by applying two-point form.
Best Answer
Draw a picture. Let our triangle be $ABC$, with right angle at $C$. Let the legs be $a$ and $b$. Let the median to side $a$ have length $13$. By the Pythagorean Theorem, we have $\frac{a^2}{4}+b^2=13^2$. Similarly, $a^2+\frac{b^2}{4}=19^2$. Solve.
Or better, don't solve. Add. We get $\frac{5}{4}(a^2+b^2)=13^2+19^2$.