Question

in the given figure ABCD is quadrilateral in which AB=AD and BC=DC prove that AC is the bisector of angle BAD and angle BCD​

Answer 1

Answer:

Consider △ ABC and △ ADC It is given that AB = AD and BC = DC AC is common i.e.

AC = AC By SSS congruence criterion △ ABC ≅ △ ADC ……… (1)

∠ BAC = ∠ DAC (c. p. c. t) So we get ∠ BAE = ∠ DAE

We know that AC bisects the ∠ BAD i.e. ∠ A

So we get ∠ BCA = ∠ DCA (c. p. c. t) It can be written as

∠ BCE = ∠ DCE So we know that AC bisects ∠ BCD i.e. ∠ C

(ii) Consider △ ABE and △ ADE It is given that AB = AD AE is common i.e.

AE = AE By SAS congruence criterion

△ABE ≅ ∠ ADE BE = DE (c. p. c. t)

(iii) We know that △ ABC ≅ △ ADC Therefore, by c. p. c. t ∠ ABC