Question

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

Answer 1

ANSWER:
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
∴ ∠ABC = ∠AD​ C
(Proven above)
(By CPCT)