Question

Let ABCD be a quadrilateral with ∠CBD = 2 ∠ADB, 

∠ABD = 2 ∠CDB and AB = CB. Then which of the following statement hold true.

I AD = CD 

II AD = BD 

III BEDF is a parallelogram 

IV AB = CD 

(a) I and III only 

(b) All of above 

(c) I, II and IV 

(d) I, III and IV

Answer 1

(a) I and III only

Given, ABCD is a quadrilateral in which, 

(1) AB = BC 

(2) ∠ABD = 2∠BDC 

(3) ∠DBC = 2∠ADB 

Let BE and BF be the bisectors of ∠ABD and ∠CBD respectively. 

Then, ∠ABE = ∠EBD = ∠BDF = y and 

∠CBF = ∠FBD = ∠BDE = x 

Now, ∠EDB = ∠FBD and ∠EBD = ∠BDF 

⇒ alternate angles are equal ⇒ ED || BF and BE || FD 

⇒ BEDF is a parallelogram. 

Let EF and BD cut each other at P and AC and BD cut each other at Q 

∴ diagonals of a parallelogram bisect each other, EP = PF 

Now by the bisector theoram

\(\frac{AE}{ED}\) = \(\frac{AB}{BD}\) = \(\frac{BC}{BD}\) = \(\frac{CF}{FD}\)                  (∵ AB = BC)

⇒ \(\frac{AE}{ED}\) = \(\frac{CF}{FD}\)

⇒ By the converse of Basic proportionality theoram, 

EF || AC 

∴ EP = PF ⇒ AQ = QC

∴ \(\frac{AB}{AQ}\) = \(\frac{BC}{QC}\) 

⇒ BQ bisects ∠ABC ⇒ x = y 

⇒ ΔABD ≅ ΔBCD ⇒ AD = CD.