In the triangle ABC, the angle BAC = 90 ° AB = AC, take a point D in the triangle to make the angle abd = 30 ° and BD = Ba, AE perpendicular to BD and e to find the bisector angle EAC of AD

In the triangle ABC, the angle BAC = 90 ° AB = AC, take a point D in the triangle to make the angle abd = 30 ° and BD = Ba, AE perpendicular to BD and e to find the bisector angle EAC of AD

BD = Ba, then the angle bad is equal to the angle BDA, and the angle abd is equal to 30 °, so the angle bad is equal to the angle BDA is equal to (180 ° -- 30 °) / 2 = 75 °, so the angle DAC = angle BAC -- angle bad = 90 ° - 75 ° = 15 °. AE is perpendicular to BD, then the angle Bae is equal to 180 ° -- 90 ° -- angle abd = 60 °, and the angle ead is equal to the angle bad -- angle BAE = 75 ° - 60 ° = 15 °, so the angle ead is equal to the angle DAC, that is, ad bisecting angle EAC

In the triangle ABC, ab = AC, take a point e above AC, take a point D on the extension line of Ba, make ad = AE, connect de and extend AB to F, find DF ⊥ BC

Just prove that the angle EFC is 90 degrees
Because angle ABC = angle ACB, angle ade = angle AED = angle CEF
Angle DAE = 2 angle ABC
So the angle CEF = 90 degrees - angle ACF, that is, the angle EFC is 90 degrees