Students in the math interest group of social activities class discussed the problem of an angle mon = 90.
(1) Question 1: Proof: As shown in figure 1, ∫△Abe and △AB'E are symmetrical about AE, ∴ AEB = ∠ AEB', BE=B'E, ∫. ∴∠ Bab ′ =∠ DCD ′, ∫∠d =∠d'= 90, ∴∠ D ′ FD+∠ D ′ CD =180, ∠. (2) Extended exploration: The conclusion found by the students in the practice group is correct. It is proved that bb'⊥AE is connected as shown in Figure 3. ∴∠ Mao +∠ AMO = 90, ∠ OBB'+∠ BME = 90, ∠AMO=∠BME, Mao =∞. ∴ MB = DN。 ∴ OM = ON, ∴ om = ob ′ = on = od ′, ∴ quadrilateral MB ′ nd ′ is a rectangle, ∴AC⊥BD, ∴ quadrilateral MB ′ nd ′ is a square.