In the given figure,
is a quadrilateral in which
and
Prove that
(i)
bisects
and ![]()
(ii) ![]()
(iii) ![]()

Given: In ABCD,
and ![]()
To prove: (i)
bisects
and ![]()
(ii) ![]()
(iii) ![]()
Proof:
(i) In ∆ABC and ∆ADC, we have,
AB = AD …given
BC = DC …given
AC = AC … common side
Hence, by SSS congruence rule,
∆ABC ≅ ∆ADC
∴ ∠BAC = ∠DAC and ∠BCA = ∠DCA …By cpct
Thus, AC bisects ∠A and ∠ C.
(ii) Now, in ∆ABE and ∆ADE, we have,
AB = AD …given
∠BAE = ∠DAE …from i
AE = AE …common side
Hence, by SAS congruence rule,
∆ABE ≅ ∆ADE
∴ BE = DE …by cpct
(iii) ∆ABC ≅ ∆ADC from ii
∴ ∠ABC = ∠ADC …by cpct
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