The quadrilateral ABCD and the quadrilateral aefg are both rhombic. The point E is on the ad side and the point G is on the extension line of the Ba side As shown in Figure 1, BD and CE are their diagonals. The extension line of Ge intersects BD at point Q and DC at point M. verify: GM ⊥ BD

The quadrilateral ABCD and the quadrilateral aefg are both rhombic. The point E is on the ad side and the point G is on the extension line of the Ba side As shown in Figure 1, BD and CE are their diagonals. The extension line of Ge intersects BD at point Q and DC at point M. verify: GM ⊥ BD

prove:
In diamond ABCD, ∠ abd = ∠ ABC / 2, ad ‖ BC,
In rhombic aefg, ∠ GAF = ∠ gae / 2,
So, gae = ABC,
Therefore, GAF = abd
So AF ‖ BD
In diamond aefg, AF ⊥ Ge,
So GM ⊥ BD

As shown in the figure, ad ‖ BC, BD, AC, quadrilateral ABCD must be diamond? If so, please explain the reason

∵AD‖BC,
The intersection of BD and AC is at point o
∠AOD＝∠COB
OA=OC
∴ΔAOD≌ ΔCOB
∴AD＝BC
The quadrilateral ABCD is a parallelogram,
Od = 0b OA = OC
The parallelogram ABCD is a diamond

As shown in the figure, the quadrilateral ABCD, cdef and efgh are all square. (1) is △ ACF similar to △ GCA? Talk about your reasons; (2) find the degree of ∠ 1 + 2

(1) Reasons: let the side length of a square be a, AC = A2 + A2 = 2A, ∵ ACCF = 2AA = 2, CGAC = 2a2a = 2, ∵ ACCF = CGAC, ∵ - ACF = ∵ ACF, ∵ ACF ∽ GCA; (2) ∵ ACF ∽ GCA,