Question

In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that
(i) AC bisects ∠A and ∠C,
(ii) BE = DE,
(iii) ∠ABC = ∠ADC.

Answer 1

​Given: ABCD is a quadrilateral in which AB = AD and BC = DC
(i)
In ∆ABC and ∆ADC, we have:
AB = AD (Given)

BC = DC (Given)
AC is common.
i.e., ∆ABC ≅ ∆ADC (SSS congruence rule)

∴ ∠BAC = ∠DAC and ∠BCA = ∠D​CA (By CPCT)
Thus, AC bisects ∠A and ∠ C.

(ii)
Now, in ∆ABE and ∆ADE, we have:
AB = AD (Given)​
∠BAE = ∠DAE​ (Proven above)
AE is common.
∴ ∆ABE ≅ ∆ADE (SAS congruence rule)
⇒ BE = DE (By CPCT)

(iii) ∆ABC ≅ ∆ADC (Proven above)
∴ ∠ABC = ∠AD​C (By CPCT)​