So ∠5=∠3=∠ 1+∠4 and ∠4=∠ 1=∠5=∠6.

So ∠5=∠3=2∠6.

Because ∠ 5+∠ 3+∠ 6 = 180.

So ∠ 5 = 72, ∠ 6 = 36.

So ∠ 4 = ∠ 6 = 36

So ∠ BAC = 108.

Wait a minute. I am writing a process. It may be a little slow.

2. Because AB=AC

So ∠ b = ∠ c.

Because AE⊥BC CD⊥AB

So ∠ AEB =∠ AEC =∠ CDB = 90.

Because < b = < c

So < 1 = < 2.

Because ∠B=∠B, ∠BDC=∠AEB

So < 1 = < 3.

Because ∠ 1=∠2

So ∠ BAC = ∠1+∠ 2 = 2 ∠1= 2 ∠ 3.

three

1) connection DE

Then in Rt△ABD, DE is the center line on the hypotenuse,

De = Bei =DC

So delta △EDC is an isosceles triangle.

So DG⊥EC

So g is the midpoint of CE (three lines are one)

so DE=BE

So ∠B=∠EDB

∠EDB=∠ECD+∠CED=2∠ECD

So ∠B=2∠BCE

4. take a little e from BC to make Be = BD.

So △BDE is an isosceles triangle.

Because in the isosceles triangle, AB = AC, ∠ A = 100, BD bisects ∠ AB=AC.

So ∠ ABC =∠ ACB = 40.

So: ∠ Abd = ∠ DBE = 20.

So in △BDE, ∠ BDE = ∠ Bed = 80.

Because ∠ bed = ∠ ECD+∠ EDC

So ∠ EDC = 40.

So de = EC

Crossing D, DG⊥BC is in G, DF⊥AB extension line is in F.

So df = DG,

Because ∠ A = 100, ∠ FAD = 80.

So RT△ADF≌RT△EDG

So de = ad

So ad = ce

So AD+BD=BC

5.

PD+PE equals CF

Pass point p as PG⊥CF in g.

So PG∨AB

So ∠ ABC = ∠ GPC, PD = FG.

Because ab = AC

So ∠ ABC = ∠ ACB

So RT△PGC≌RT△CEP

So PE = CG

So PD+PE = cf