In the quadrilateral ABCD, AB is vertical to BC, DC is vertical to BC, P is a point on ad, PA = AB, PD = CD, then how many degrees is the angle BPC

In the quadrilateral ABCD, AB is vertical to BC, DC is vertical to BC, P is a point on ad, PA = AB, PD = CD, then how many degrees is the angle BPC

∵AB⊥BC,CD⊥BC
∴AB‖CD
∴∠A+∠D=180°
∵AP=AB
∴∠APB=90°-1/2∠A
∵PD =CD
∴∠CPD=90°-1/2∠D
∴∠BPC=180°-∠APB-∠CPD=180°-(90°-1/2∠A)-(90°-1/2∠D)=1/2(∠A+∠D)=90°

As shown in the figure, in ladder ABCD, ad ‖ BC, ab = DC, P is a point outside the ladder ABCD, PA and PD intersect BC at points E and f respectively, and PA = PD. (1) write three pairs of triangles that you think are congruent in the graph (no auxiliary lines are added); (2) select any pair of congruent triangles that you write in (1) to prove

(1) (1) △ ABP ≌ △ DCP; (2) △ Abe ≌ △ DCF; (3) △ BEP ≌ △ CFP; (4) △ BFP ≌ △ CEP; (2) the reference answers to △ ABP ≌ △ DCP are given below. It is proved that: ∫ ad ≌ BC, ab = DC, ≌ trapezoid ABCD is isosceles trapezoid; ∫ bad = ≌ - CDA; and ∫ PA = PD, ≌ - pad = ≌ - PDA, ≌ - bad -

The midpoints of the edges AB, BC, CD and Da of the space quadrilateral ABCD are e, F, G and h respectively
If the length of BD perpendicular to AC, tangent to BD and AC is 2 and 4 respectively, then eg & # 178; + HF squared value

Because e, F, G and H are the midpoint of AB, BC, CD and Da respectively
So EF, GH parallel AC, FG, eh parallel BD, and EF = GH = AC / 2 = 2, FG = eh = BD / 2 = 1
And because BD is vertical to AC
So efgh is rectangular
So eg ^ 2 + HF ^ 2 = 2 * (EF ^ 2 + FG ^ 2) = 10
So the answer is: 10