RT, in triangle ABC, ab = AC = 10, point D on BC, de parallel AC, DF parallel AB, find the perimeter of quadrilateral AEDF

RT, in triangle ABC, ab = AC = 10, point D on BC, de parallel AC, DF parallel AB, find the perimeter of quadrilateral AEDF


The perimeter is ab + AC = 20
Because AB = AC, so ∠ B = ∠ C, and because De is parallel to AC, DF is parallel to AB, then ∠ B = ∠ EDB, ∠ C = ∠ FDC, then EB = ed, FC = FD, the perimeter of quadrilateral is equal to AE + AF + de + DF = AE + be + AF + FC = AB + AC = 20



As shown in the figure, in △ ABC, ab = AC = 5, D is the point on the edge of BC, de ∥ AB intersects AC at point E, DF ∥ AC intersects AB at point F, then the perimeter of quad afde is______ .


∵ AB = AC = 5, ∵ B = ∵ C, from DF ∥ AC, ∵ FDB = ∵ C = ∵ B, ∵ FD = FB, similarly, de = EC. ∵ perimeter of quadrilateral afde = AF + AE + FD + de = AF + FB + AE + EC = AB + AC = 5 + 5 = 10



As shown in the figure, in △ ABC, de ‖ BC, and s △ ade: s quadrilateral bced = 1:2, BC = 26. Find the length of de


∵ s △ ade: s quadrilateral bced = 1:2, s △ ABC = s △ ade + s quadrilateral DBCE, ∵ s △ ade: s △ ABC = 1:3, and ∵ de ‖ BC, ∵ ade ∵ ABC, ∵ s △ ade: s △ ABC = (DEBC) 2, and ∵ BC = 26, ∵ de = 22



As shown in the figure, in △ ABC, de ‖ BC, and s △ ade: s quadrilateral bced = 1:2, BC = 26. Find the length of de


∵ s △ ade: s quadrilateral bced = 1:2, s △ ABC = s △ ade + s quadrilateral DBCE, ∵ s △ ade: s △ ABC = 1:3, and ∵ de ‖ BC, ∵ ade ∵ ABC, ∵ s △ ade: s △ ABC = (DEBC) 2, and ∵ BC = 26, ∵ de = 22



As shown in the figure, in isosceles △ ABC, ab = AC, ∠ BAC = 100 °, extend AB to D, make ad = BC, connect DC, then the degree of ∠ BCD is______ .





As shown in the figure, in isosceles △ ABC, ab = AC, ∠ BAC = 100 °, extend AB to D, make ad = BC, connect DC, then the degree of ∠ BCD is______ .





Known: as shown in the figure, in △ ABC, ab = AC, ad bisects ∠ BAC, CE ⊥ AB in E, intersects ad in F, AF = 2CD, find the degree of ∠ ace


∵ AB = AC, ad bisection ∠ BAC, ∵ BD = DC, ad ⊥ BC, that is BC = 2CD, ∵ AF = 2CD, ∵ AF = BC, ∵ CE ⊥ AB, ad ⊥ BC, ∵ AEF = ∠ BEC = ∠ ADC = 90 °, ∵ AFE = ∠ DFC, ∵ AEF + ∠ AFE + ∠ EAF = 180 °, DFC + ∠ FDC + ∠ FCD = 180 °, and ∵ EAF = ∠ FCD, in △ AEF



In the triangle ABC, angle a = 70 degrees. If angle B intersects the bisector of angle c at point E, then angle BEC =?


If the bisector of angle B and angle c intersects at point E, then point E is a triangle ABC inscribed circle,
The angle BEC is the center angle
Angle BEC = 1 / 2 (angle ACB + angle ABC)
Because angle ACB + angle ABC + angle a = 180 degrees
So: angle BEC = 1 / 2 (angle ACB + angle ABC) = 1 / 2 (180 angle a) = 1 / 2 (180-70) = 55 degrees



Known triangle ABC, angle c is equal to 90 degrees, be is the angle bisector of angle B, intersection AC at F, find the degree of angle BEC, urgent!


The problem is serious. Let's hand it over to AC and E··



In △ ABC, if the diplomatic bisector of ∠ B and the outer bisector of ∠ C intersect at point E, then ∠ BEC is equal to ()
A 1/2(90°-∠A) B 90°-∠A C 1/2(180°-∠A) D 180°-∠A


The sum of the external angles of ∠ B + C = 180 ° - a ∠ B ∠ C is 360 ° - (180 ° - a) = 180 °+ ∠ a ∠ EBC + ∠ ECB = (180 °+ ∠ a) △ 2 = 90 ° + 1 / 2 ∠ a ∠ BEC = 180 ° - (90 ° + 1 / 2 ∠ a) = 90 ° - 1 / 2 ∠ a = 1 / 2 (180 ° - a)