As shown in the figure, ∠ a = ∠ B, CE ‖ Da, CE intersects AB with E

As shown in the figure, ∠ a = ∠ B, CE ‖ Da, CE intersects AB with E


It is proved that ∵ CE ∥ Da, ∵ a = ∵ CEB. And ∵ a = ∵ B, ∵ CEB = ∵ B. ∵ CE = CB. ∵ CEB is isosceles triangle



As shown in the figure, ∠ a = ∠ B, CE ‖ Da, CE intersects AB with E


It is proved that ∵ CE ∥ Da, ∵ a = ∵ CEB. And ∵ a = ∵ B, ∵ CEB = ∵ B. ∵ CE = CB. ∵ CEB is isosceles triangle



Known: as shown in the figure, Da ⊥ AC, EC ⊥ Ca, point B on AC, DB ⊥ be, ab = CE


It is proved that: if ∵ Da ⊥ AC, EC ⊥ Ca ≁ a = ≁ C = 90 °, e + ≁ CBE = 90 ° DB ⊥ be ≁ DBE = 90 ° then abd + ⊥ CBE = 90 °, abd = ≌ e, in △ abd and △ CEB ≌ a = ≌ C (proved) ∠ abd = ≌ e (proved) AB = Ce (known) ≌ abd ≌ CEB (ASA)



&As shown in the figure, ∠ a = ∠ B = 60 ° CE ‖ Da, CE intersects AB with E


It is proved that: ∵ a = 60 °, CE ∥ Da, ∵ CEB = 60 °, ∵ a = ∵ B = 60 °, ∵ CEB = ∵ B = ∥ ECB = 60 ° and ∫ CEB is an equilateral triangle