In △ ABC, ∠ C = 90 °, ab = 13, BC = 5, then the radius of its circumscribed circle is______ The radius of the inscribed circle is______ ．

In △ ABC, ∠ C = 90 °, ab = 13, BC = 5, then the radius of its circumscribed circle is______ The radius of the inscribed circle is______ ．

∵ in △ ABC, ∠ C = 90 °, AB is the diameter of its circumscribed circle, ∵ AB = 13, ∵ the radius of its circumscribed circle is 6.5; ∵ in △ ABC, ∵ C = 90 °, ab = 13, BC = 5, ∵ AC = AB2 − BC2 = 12, let the radius of the inscribed circle be r, ∵ s △ ABC = 12ac · BC = 12 (AB + AC + BC) r, ∵ r = AC · bcab + AC + BC = 12 × 512 + 13 + 5 = 2, and the radius of the inscribed circle be 2

In the isosceles RT triangle ABC, ab = AC, angle BAC = 90 degrees, be bisection angle BAC intersects AC at e, CD is made through C, be is perpendicular to D, and ad is connected
Write in congruence and equivalency,

Because of the angle BAC = 90 degrees, CD ⊥ BD, the four points of a.d.c.b are in a circle ∠ DAC = ∠ DBC, because AC = AB, the triangle cab is an equilateral triangle, ∠ CBA = 45 ° and because BD is the angle bisector of ∠ CBA, ∠ DBC = ∠ DBA = 22.5 °, DAC = 22.5 °∠ AEB = ∠ EAB = 90 ° - 22.5 °= 67.5 °∠ ADB =

As shown in the figure, in the RT triangle ABC, the angle c is equal to 90 ° and the angle B is equal to 30 ° ad is the bisector of the angle BAC, and the de bisector angle ADB intersects AB with E
Prove that triangle ACD is equal to triangle bed
This is the 14th question on page 3 of the golden key of mathematics published by people's education press

It is proved that in ∵ RT triangle ABC, angle c is equal to 90 ° and angle B is equal to 30 °
∴AB=2AC
∵∠B=∠BAD
∴AD=BD
∴BE=AE=AC
The angle c equals 90 degrees
∴∠CDA=60
∴∠BDE=60=∠CDA
∵∠B＝∠DAC
The triangle ACD is equal to the triangle bed

As shown in the figure, in RT △ ABC, ∠ C = 90 °, ad bisects ∠ BAC, BD bisects ∠ CBE, then ∠ ADB=______ Degree

Suppose ∠ cab = x °, ad bisects ∠ BAC, ∠ DAB = 12 ∠ cab = x ° 2, then in RT △ ABC, ∠ ABC = 180 °～ ACB - ∠ cab = 180 ° - 90 ° - x ° = 90 ° - x °, ∫ CBE is the outer angle of RT △ ABC, ∫ CBE = ∠ cab + ∠ ACB = x ° + 90 °, ∫ CBD = 12 ∠ CBE = 12 (x ° + 90 °)