Known: as shown in the figure, in the isosceles subangular triangle ABC, D is the midpoint of the hypotenuse AB, P is on BD, PM is perpendicular to BC, m, PN is perpendicular to AC, N, DM is equal to DN

Known: as shown in the figure, in the isosceles subangular triangle ABC, D is the midpoint of the hypotenuse AB, P is on BD, PM is perpendicular to BC, m, PN is perpendicular to AC, N, DM is equal to DN

Proof: connect CD
∵AC＝BC,∠ACB＝90
∴∠A＝∠B＝45
∵PM⊥BC,PN⊥AC
Ψ rectangular pmcn
∴PM＝CN
And ∵ PM ⊥ BC, ∠ B = 45
∴BM＝PM
∴BM＝CN
∵CM＝BC-BM,AN＝AC-CN
∴CM＝AN
∵ D is the midpoint of ab
∴CD＝AD＝BD,∠ACD＝∠BCD＝∠ACB/2＝45
∴∠BCD＝∠A
∴△ADN≌△CDM （SAS）
∴DM＝DN

The length of BC at the bottom of ABC is 2 and 3, ab = AC, and the vertex angle is 120 ° to find the diameter of its circumscribed circle

Make ad ⊥ BC at D, lengthen AD and circumscribed circle at E
Then ∠ bad = 1 / 2 ∠ BAC = 60 ° BD = 1 / 2 BC = √ 3
∴AD :BD:AB=1：√3:2
And BD = 1 / 2 BC = √ 3
∴AB=2
∵ AE vertical bisection BC
The AE is the diameter
∴∠ABE=90° ∠E=90°- ∠BAE=30°
∴AE=2AB=4

If the waist length of the isosceles triangle ABC is 3cm and the vertex angle is 120 °, the diameter of its circumscribed circle is

The diameter of circumscribed circle is 3 × 2 = 6 cm