The triangle ABC is an isosceles right triangle, the angle BAC is equal to 90 degrees, be is the bisector of the angle, ED is perpendicular to BC. If ad = 4, EB = 5, find the face of quad ABDE

Be is the angular bisector
So the angle Abe = EBD has ∠ EAB = EDB = 90, so △ AEB is equal to EBD
So the area of quad ABDE is 5 * 4 / 2 = 10

In the isosceles triangle ABC, ∠ C = 90 degrees, AC = BC, ad bisection ∠ BAC, be perpendicular to e, prove be = 1 / 2ad

The title is incomplete. "Be perpendicular to e" should be "be perpendicular to e"
prove:
Extend be and AC to f
Because ad bisects cab, AE ⊥ be
Therefore, BAE = FAE and bea = FEA
And because AE = AE
So △ BAE ≌ △ FAE (ASA)
So be = Fe
So BF = 2be
Because ∠ CBF ＋ f = 90 degrees, ∠ FAE ＋ f = 90 degrees
Therefore, CBF = FAE = CAD
And because BC = AC, ∠ BCF = ∠ ACD = 90
So △ BCF ≌ △ ACD (ASA)
So BF = ad
So ad = 2be
That is be = ad / 2
For reference! Jswyc

It is known that: as shown in the figure, D is the midpoint of BC, ED is vertical AB, FD is vertical AC, be = CF in right triangle ABC
Line.

Certification:
In ∵ △ bed and △ CFD, BD = CD, be = CF, ∠ bed = ∠ CFD
In the case of ≌ △ bed ≌ △ CFD, there is ≌ B = ≌ C
In ∵ abd and ∵ ACD, BD = CD, ad = ad, ∠ B = ∠ C
If ≌ △ abd ≌ △ ACD, then ∠ bad = ∠ CAD
Then ad is the angular bisector of the right triangle ABC

It is known that: as shown in the figure, in △ ABC, the bisector am of ∠ C = 90 °, BAC = 60 ° is 15cm, and the length of BC is calculated

∵ am is the bisector of ∵ BAC, ∵ BAC = 60 °, ∵ MAC = 30 °, ∵ MC = 12am = 7.5cm, ∵ AC = am2-mc2 = 152-7.52 = 1523, ∵ in △ ABC, ∵ C = 90 °, ∵ BAC = 60 °, ∵ ABC = 30 °, ∵ AB = 2Ac = 153, ∵ BC = ab2-ac2 = (153) 2 - (1523) & nbsp; 2 = 452

The triangle ABC is an isosceles right triangle, the angle BAC is equal to 90 degrees, be is the bisector of the angle, ED is perpendicular to BC. If ad = 4, EB = 5, find the face of quad ABDE