In △abc, ad and be are the angle bisectors of ∠a and ∠b and de ║ ab . find the length of de if perimeter of abde is 30 cm and ab = 12 cm

DE = 6 cm

Step-by-step explanation:

Let DE = x cm.

Since DE is parallel to AB therefore by the alternate interior angles theorem, m∠BAD = m∠ADE and m∠ABE = m∠DEB ............(1)

As AD is an angle bisector of ∠A, therefore m∠EAD = m∠DAB ; Since BE is an angle bisector of ∠B ⇒ m∠ABE = m∠EBD.

Therefore, from (1) We get ,  m∠EAD = m∠ADE and m∠EBD = m∠BED.

So, the triangles ADE and EDB are then isosceles with AE = ED and ED = DB.

So AE = DE = DB = x, and since the perimeter of ABDE is 30 cm, then

12 + x + x + x = 30

⇒ 12 + 3x = 30

⇒ x = 6

Hence, the length of DE is 6 cm.

DE=6cm

Step-by-step explanation:

Let x=DE, since AB║DE, therefore ∠BAD=∠ADE and ∠ABE=∠BED.        (1)

Also, we are given that AD is the angle bisector of ∠A and BE is the angle bisector of ∠B,

Therefore, ∠EAD=∠DAB and ∠ABE=∠EBD                                             (2)

From (1) and (2), ∠EAD=∠ADE and ∠EBD=∠BED

⇒The ΔADE and ΔEDB then becomes the isosceles triangle with AE=Ed and ED=DB (Sides opposite to equal angles are always equal)

Therefore, AE=DE=DB=x

We are given that the perimeter of ABDE is 30 cm, therefore,

Perimeter of ABDE= sum of all the sides of ABDE

⇒30=AB+BD+DE+AE

⇒30=12+3x

⇒30-12=3x

⇒x=6cm

Therefore, the length of DE= 6cm

that is the sum

step-by-step explanation:

there are no other answers without knowing the variables values