Question

In Pentagon ABCDE show that AB+AE+BC+CD+ED=2AD

Answer 1

Answer :

Given In a pentagon ABCDE , AB = AE and BC = ED and ∠ ABC = ∠ AED , And we join AC and AD , and form our diagram , As :



i ) Now In ∆ ABC and ∆ AED

AB = AE ( Given )

∠ ABC = ∠ AED ( Given )

And

BC = ED ( Given )

Hence

∆ ABC ≅∆ AED ( By SAS rule )

So,

AC = AD --------------------( 1 ) ( by CPCT ) ( Hence proved )

And

∠ BCA = ∠ EDA --------------------( 2 ) ( by CPCT )

ii ) Now In ∆ CAD , We know

AC = AD ( from equation 1 )

So , from base angle theorem , we know

∠ ACD = ∠ ADC ---------- ( 3 )

And

∠ BCD = ∠ BCA + ∠ ACD ---------- ( 4 )
And
∠ EDC = ∠ EDA + ∠ ADC , So from equation 2 and 3 , we get

∠ EDC = ∠ BCA + ∠ACD
From equation 4and5