Solution: The two triangles are equal in height, but different in base, so the ratio of base is the ratio of area.

The passing point D is df⊥ac de⊥ab, and the vertical feet are E and F respectively.

∵AD is the bisector of ∠BAC, DE⊥AB, DF⊥AC,

AD = AD

∴△ADE≌△ADF

∴DE=DF，

∴S△ABD= 1/2DE？ ab blood type

∴S△ACD= 1/2？ DF？ Alternating current

∴S△ABD:S△ACD= 1/2DE？ AB: 1/2？ DF？ AC=AB:AC=5:3

2、

Solution: Find the perimeter of △DBE, that is, find the value of DE+EB+BD.

∫ad ∠ ∠CAB, ∠ c = 90, DE⊥AB.

∴DC=DE.

Prove that △ ACD △ aed. ∴ AC = AE。

AC = BC，

∴de+eb+bd=dc+eb+bd=bc+eb=ac+eb=ae+eb=ab

∫AB = 10cm，

∴△ perimeter of ∴△DBE = db+be+de =10cm.

∴△ The circumference of ∴△DBE is10cm. ..

3、

It is proved that the intersection D is DE⊥AB, the extension line of intersection BA is E, and DF⊥BC is F.

∵DE⊥AB, DF⊥BC and BD share ABC equally.

∴DE=DF？ (Angular bisector property), ∠ AED = ∠ CFD = 90

∵∠DAB+∠DAE= 180，？ ∠DAB+∠DCB= 180

∴∠DAE=∠DCB

∴△ADE≌△CDF？ (AAS)

∴AD=CD

4、

Make EF perpendicular to AD and point to f through e.

Bisection ADC

∴∠CDE=∠EDF,DE=EF=EB

∠∠AFE =∠ABE = right angle, AE=EA.

∴△ABE≌△AEF

∴∠EAF=∠EAB

That is, AE divides ∠BAD.

It is not easy. I hope I can help you.