Solution 1.
Analysis:
Let ABC be the triangle with BC = a, B-C, BB' + CC' = h_b + h_c given.
Let D be the point on CA such that AD = AB with the A between D,C
(DC = DA + AC = AB + AC).
The triangle ABD is isosceles with ang(ADB) = ang(ABD) = A/2.
Let E be the orthogonal projection of C on DB.
In the triangle EBC we have:
ang(CEB) = 90 d., BC = a, ang(EBC) = ang(BCD + BDC) = C + (A/2) =
90 - ((B-C)/2) d.
Therefore EC is known, since the triangle EBC can be constructed.
From the similar triangles ECD and B'BD we have:
EC / DC = BB' / BD or
EC / (AB + AC) = BB' / DB (1)
From the similar right triangles ABB', ACC' we have:
BB' / AB = CC' / AC ==>
BB' / AB = CC' / AC = (BB' + CC') / (AB+AC) ==>
BB' = AB(BB' + CC') / (AB + AC) (2)
From (1) and (2) we get:
EC / (BB' + CC') = AB / DB : fixed,
since EC and BB' + CC' are known.
Therefore the triangle ADB "remains similar to itself", that is, it has known angles. Therefore angle A is known, since ang(BDA) = ang(ABD) = A/2
Construction: We leave it to the reader.
Solution 2.
Analysis:
Let h_a = AD, AE be the altitude, angle bisector from A, resp.
The parallel from E to AB intersects AC at Z.
The triangle ZAE is isosceles with EZ = AZ
(since angles ZEA = BAE = EAZ = A/2)
EZ / AB = CZ / CA = (CA - AZ) / CA = (CA - EZ) / CA
==> 1/EZ = 1/b + 1/c (1)
2*area(ABC) = ah_a = bh_b = ch_c ==>
h_2 + h_3 = ah_a (1/b + 1/c) (2)
From (1) and (2) we get that
h_a / EZ = AD / EZ = (h_b + h_c) / a : fixed (3)
The triangle DAE has known angles:
ang(ADE) = 90 d., ang(DAE) = (B-C)/2, (DEA) = 90 - ((B-C)/2)
Therefore AD / AE = h_a / AE is fixed. (4)
From (3) and (4) by division we get that:
AE / EZ is fixed.
Now, in the isosceles triangle ZAE we have that AE / EZ is fixed, therefore "it remains similar to itself", that is its angles are known. So A = 2*ang(EAZ) is known.
Construction: We leave it to the reader.
Problem for the reader:
To construct ABC if are given a, B-C, h_c - h_b.
Δεν υπάρχουν σχόλια:
Δημοσίευση σχολίου