In △ ABC, ∠ ABC = 30 ° with BC and ab as sides, make equilateral △ BCD and equilateral △ ace to connect be Verification: ab & sup2; + BC & sup2; = be & sup2;

It is impossible to make equilateral △ ace with ab as the edge!

As shown in the figure, in the known triangle ABC, take AB and AC as equilateral images, make equilateral triangle ABF and equilateral triangle ace, connect be, CF, and intersect at point D,
Proof: ad bisector EDF

As shown in the figure, it is proved that ∠ 1 = 2
∵ equilateral △ ABF and equilateral △ ace
AF=AB,AC=AE,∠FAB=∠EAC=60°
∴∠FAB+∠BAC=∠EAC+∠BAC
That is, FAC = BAE
∴△FAC≌△BAE
The area of FC = be, △ fac = △ BAE
(the original picture is too small, another one has been drawn.)
The two triangles have the same area and the same base
When passing point a, make high Ag and ah respectively
We can get high equality, that is Ag = ah
The point with equal distance from the inside of the angle to both sides of the angle is on the bisector of the angle

Taking AB and AC of the triangle AB as the sides, making equilateral triangles ABC and ACE outside the triangle, connecting CD and be to intersect o, we prove that OA bisection ∠ doe

It is proved that: am ⊥ be at m, an ⊥ CD at n ∵ equilateral ⊥ abd, equilateral ⊥ ace ≌ AB = ad, AC = AE, ∠ bad ≌ CAE = 60 ⊥ BAE ⊥ CAE ⊥ BAC, ∠ DAC ≌ bad ≌ BAC ≌ BAE ≌ be = CD, s △ Abe = s △ ADC ⊥ am ⊥ be, an ⊥ CD

Taking AB and AC of Δ ABC as sides, make isosceles right angles △ abd and isosceles right angles △ ace outward respectively, connect be and CD at o point

(unless AB = AC, and is bisecting the angle BAC, it may not be possible to do it, because the angle doa is not necessarily 90) from the question, DB = EC, DBC = angle ECB, be = be, so triangle DBC is equal to triangle ECB, so angle EBC = angle DCB, so ob = OC, because Ao = Ao, ab = AC, so triangle AOB is equal to triangle AOC

In △ ABC, ∠ ABC = 30 ° with BC and ab as sides, make equilateral △ BCD and equilateral △ ace to connect be Verification: ab & sup2; + BC & sup2; = be & sup2;