In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC.
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.

 

Solution:

Given:  ABCD is a quadrilateral in which AB = AD and BC = DC  
(i)
In ∆ABC and ∆ADCwe 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 ∆ADEwe 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)​

 

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