As shown in the figure, in △ ABC, be bisects ∠ ABC intersects AC at point E, de ‖ BC intersects AB at point D, ∠ ade = 70 ° and calculates the degree of ∠ DEB Don't use the formula (one outer angle of a triangle is equal to the sum of two nonadjacent outer angles),

As shown in the figure, in △ ABC, be bisects ∠ ABC intersects AC at point E, de ‖ BC intersects AB at point D, ∠ ade = 70 ° and calculates the degree of ∠ DEB Don't use the formula (one outer angle of a triangle is equal to the sum of two nonadjacent outer angles),

Because de / / BC
therefore

As shown in the figure, it is known that be bisects ∠ ABC, ∠ DEB = ∠ DBE, then are ∠ ade and ∠ ABC equal? Why?

The reasons are as follows: because be bisects ∠ ABC (known), so ∠ CBE = ∠ DBE (meaning of angle bisector). Because ∠ DEB = ∠ DBE (known), so ∠ DEB = ∠ CBE (equivalent substitution), so de ‖ BC (equal internal angle, parallel two lines), so ∠ ade = ∠ ABC (parallel two lines, equal angle)

As shown in the figure, in triangle ABC, be bisects angle ABC, and angle DBE = angle DEB, indicating that De is parallel to BC

Be bisector angle ABC
So angle DBE = angle EBC
Also known as angle DBE = angle DEB
So angle EBC = angle DEB
So de ‖ BC

As shown in the figure, in △ ABC, the bisectors BD and CE intersect at O. (1) ∠ abd = 25 ° and ∠ ace = 30 ° to find the degree of ∠ a
(2) The degree of DOC can be calculated by ∠ a = 80 °, abd = 25 °

∵ BD and CE are equally divided into ∵ ABC, ∵ ACB, ∵ ABC = 2 ∵ abd = 50 °, ∵ ace = 2 ∵ ace = 60 °, ∵ a = 180 ° - (∵ ABC + ∵ ACB) = 70 ° respectively

As shown in the figure, in △ ABC, the angle bisectors BD and CE intersect at O (1) ∠ abd = 25 ° and ∠ ace = 25 ° to find the degree of ∠ a
As shown in the figure, in △ ABC, the bisectors BD and CE intersect at point o
(1) The degree of ∠ a can be calculated by ∠ abd = 25 ° and ∠ ace = 25 °
(2) How to calculate the degree of doc

(1) Because BD and CE are bisectors of ∠ ABC and ∠ ACB, so ∠ ABC = 2 ∠ abd = 50 °, ACB = 2 ∠ ace = 50 °, a = 180 ° - 50 ° - 50 ° = 80 °
(2) Because BD and CE are bisectors of ∠ ABC and ∠ ACB, so ∠ ABC = 2 ∠ abd = 50 °, DBC = ∠ abd = 25 °, ACB = 180 ° - A - ∠ ABC = 50 °, BCE = 1 / 2 ∠ ACB = 25 °, Doc = ∠ DBC + ∠ BCE = 50 °

As shown in the figure, in △ ABC, BD is the angular bisector, CE is high, and ∠ ACB = 60 ° and ∠ ADB = 97 °, calculate the degree of ∠ A and ∠ ace

∫ ADB = ∠ DBC + ∠ ACB, ∫ DBC = ∠ ADB - ∠ ACB = 97 ° - 60 ° = 37 °. ∫ BD is the angular bisector, ∫ ABC = 74 °, ∫ a = 180 ° - ABC - ∠ ACB = 46 °. ∫ CE is high, ∫ AEC = 90 °, ∫ ace = 90 ° - a = 44 °

As shown in the figure, AE and CE divide ∠ BAC and ∠ ACD equally, and ∠ e = 90 °, then ab ‖ CD, is this conclusion correct? Why?

This conclusion is correct, i.e. ab ‖ CD, for the following reasons: ∵ e = 90 °, ∵ CAE + ace = 90 °, ∵ AE and CE are equally divided into ∵ BAC, ∵ ACD, ∵ CAE = ∵ BAE = 12 ∵ cab, ∵ ace =