It is known that BD is the middle line of the triangle ABC, and the perimeter of the triangle abd is 2cm longer than that of the triangle BCD. If the perimeter of the triangle ABC is 18cm and AC = 40cm, the lengths of AB and BC are obtained

It is known that BD is the middle line of the triangle ABC, and the perimeter of the triangle abd is 2cm longer than that of the triangle BCD. If the perimeter of the triangle ABC is 18cm and AC = 40cm, the lengths of AB and BC are obtained

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It is known that in triangle ABC, ad is perpendicular to BC, perpendicular foot is D, be is angle ABC bisector, EB = EC, AB + BD = CD

Make AF parallel to be and CB extension line parallel to F, because the angle EBC = ECB = Abe = BAF = AFB, so AF = AC, DF = DC, and because Ba = BF, so AB + BD = CD

It is known that ad is the height of triangle ABC, point E is on AB, EB = EC, point F is the midpoint of BC, angle 1 = angle 2, and DG is parallel to ab

It is proved that EB = EC and F is the midpoint of BC
So EF is perpendicular to BC
Because ad is the height of triangle ABC
So ad is perpendicular to BC
So ad is parallel to ef
So angle bef = angle bad
Because EF is the bisector of angle bec (three lines in one)
So the angle bef = the angle CEF
So angle bad = angle FEC
Because angle ADG = angle bef (angle 1 = angle 2, this is a guess, because there is no graph)
So angle bad = angle GDA
So dg / / AB
Missing: ∵ ∠ ADG = ∠ bef
∴∠BAD=∠GDA