It is known that in △ ABC, ab = AC, D is the midpoint of BC side, P is any point on ad, PE ⊥ AB is in E, PF ⊥ AC is in F. try to explain: (1) PE = pf; (2) Pb = PC

It is known that in △ ABC, ab = AC, D is the midpoint of BC side, P is any point on ad, PE ⊥ AB is in E, PF ⊥ AC is in F. try to explain: (1) PE = pf; (2) Pb = PC

It is proved that: (1) ∵ AB = AC, D is the midpoint of BC side, ∵ ad bisects ∵ BAC, and ∵ PE ⊥ AB in E, PF ⊥ AC in F, ∵ PE = pf; (2) ∵ AB = AC, D is the midpoint of BC side, ∵ ad vertical bisects BC, that is, ad vertical bisects BC, and ∵ P is any point on ad, ∵ Pb = PC

In △ ABC, points D, e and F are the midpoint of BC, Ca and ab respectively, then vector AB + AD + BC + be + CF=

In △ ABC, because vector Ba + AC + CB = 0, vector 1 / 2BA + 1 / 2Ac + 1 / 2CB = 0, because points D, e and F are the midpoint of BC, Ca and ab respectively, vector 1 / 2BA = BF = BC + CF1 / 2Ac = AE = AB + be1 / 2CB = CD = Ca + ad, 1 / 2BA + 1 / 2Ac + 1 / 2CB = BC + CF + AB + be + Ca + AD0 = BC + CF + AB + be + ca

Let m be an interior point of △ ABC, ∠ BAC = 30 ° and define f (m) = (m, N, P), where m, N, P are the areas of △ MBC △ MAC △ mAb respectively. If ab × AC = 4 √ 3 and f (m) = (1, N, P), then the minimum value of 1 / N + 4 / P is

Vector AB * AC = BC * cosa = BC * radical 3 / 2 = 2 radical 3 so BC = 4 so s △ ABC = 1 / 2 + X + y = 1 / 2 * BC * Sina = 1 / 2 * 4 * 1 / 2 = 1x + y = 1 / 2 so 1 / x + 4 / y = 2 * (x + y) * (1 / x + 4 / y) = 2 (1 + 4 + Y / x + 4x / y) = 2 (5 + Y / x + 4x / y) ≥ 2 (5 + 4) = 18 so the minimum value is 18

If the bisector ad of triangle ABC, a = 60, a intersects B C at point D, ab = 3 and vector ad = 1 / 3aC + MAb (M belongs to R), then the length of ad is longer

As shown in the figure, if we make DG ‖ AB, DH ‖ AC, then the vector ad = ah + AG, so Ag = 1 / 3aC, because ad bisects ∠ BAC, so ∠ bad = ∠ DAC = 30, because DG ‖ AB, so ∠ ADH = 30 = ∠ Dah, so ah = DH. Similarly, Ag = DG proves that △ ADH is all equal to △ ADG, so Ag = DH = 1 / 3aC, and because △ BDH is similar to △ BCA, so BH = 1 / 3ba = 1