P is a point in the equilateral triangle ABC, PC = 5, PA = 3, Pb = 4, find the degree of angle APB

P is a point in the equilateral triangle ABC, PC = 5, PA = 3, Pb = 4, find the degree of angle APB

Take PA as one side, make an equilateral triangle Apq outward, connect BQ, we can know that PQ = PA = 3, ∠ Apq = 60 °, because AB = AC, PA = QA, ∠ cap + ∠ PAB = 60 ° = ∠ PAB + ∠ BAQ, that is: ∠ cap = ∠ BAQ, so △ cap ≌ Δ BAQ can get: CP = BQ = 5, in △ bpq, PQ = 3, Pb = 4, BQ = 5, from Pythagorean theorem, we know that △ bpq is straight

As shown in the figure, it is known that in equilateral △ ABC, D is a point on BC, and △ DEB is an equilateral triangle. Connect CE and extend the extension line of intersection AB to point m, connect AD and extend the extension line of intersection AB to point n, and then connect Mn

ABC and debdeb are equiltriangle, ABC = AB, ABC = AB, ABC = DBE = 60 ° DB = EB, DB = EB, in △ ADB and △ CBE, ABC = AB, ABC = ABC and △ DEB are equiltriangle, ABC = AB, ABC = AB, ABC = ABC and DEB = DBE = 60 ° DB = EB, ABC = AB, ABC = AB, ABC = AB, ABC = ABC = ABC = ABC = ABC = ABC = ABC = ABC = ABC, ABC = AB = AB, ABC = ABC = AB = ABC = ABC = ABC = ABC = ABC = 60 ° DBE = DBE = 60 ° DB = 60 ° DB = EB, ABC = ABC = AB, ABC = AB = ABC = ABC = ABC, in △ ABN and △ CBN, and in △ ABN and △ CBM, the \ \ \\\ \\\\n = ∠ In addition, ∵ △ ABN ≌ △ CBM (ASA), ≌ △ BN = BM. In addition, ∵ {NBM = 180 ° -} ABC -} DBE = 60 ° and ≌ △ BMN is an equilateral triangle

Mathematics problems in grade two of junior high school (equilateral triangle)
De is on the extension line of sides AC and ab of equilateral triangle ABC respectively, and CD = AE, try to explain DB = De

Connect de
Equilateral triangle ABC
The results show that ab = AC = BC, and the three internal angles are 60 degrees
∵CD=AE
∴AB/AE=BC/CD
∴AC||DE
Ψ∠ bed = ∠ BDE = 60 degrees
The triangle EBD is an equilateral triangle
DB=DE