Question

i beg please help m eits really hard

Answer 1

Step-by-step explanation:

We know ΔBAD is isosceles ,

So ∠ABD = ∠BDA = 72°

So we get ∠DAB = 180° - (2 * 72°) = 36°

Also ∠BDC = 180° - 72° = 108°

Since we also know ΔBAC is isosceles , we have ∠DAB = ∠ACB = 36°

So we get ∠DBC = 180° - (108 + 36)° = 36°

Now we note that ∠ACB = ∠DBC

So from here we proved that ΔDBC is isosceles .

Answer 2

Answer:

given BA = BC, AB = AD , |ABD = 72°

to prove that ∆ BCD is isosceles

Step-by-step explanation:

in ∆ ABD, |ABD = |ADB = 72° [ angles opposite to equal sides are equal ]

therefore |BAD = 180° - (72° + 72°) = 36° [ angle sum property ]

now |BCD = |BAD = 36° [ AB = BC angle opposite to equal sides are equal ]...........(1)

in ∆ BCD, |BDC = 180° - |BDA = 180° - 72° = 108°

[ linear pair ]

in ∆ BCD, | DBC = 180° - (108° + 36°) = 180° - 144°

= 36° ..........(2)

therefore |BCD = |DBC = 36° [ from equation 1 & 2]

=> DB = DC [ sides opposite to equal angles are equal ]

therefore ∆ BDC is isosceles triangle