Question

A
6. In the given figure, ABCD is a quadrilateral
in which AB = AD and BC = DC. Prove that
(i) AC bisects ZA and ZC, (ii) BE = DE,
(iii) ZABC = ZADC.
B
-
D
E
C​

Answer 1

Given : In quadrilateral ABCD , AB=AD , BC=DC

1) in triangle ABC and ADC

AB = AD (given

BC= DC (given

AC=AC ( common side

triangle ABC ~ triangle ADC ( by SSS rule

angle BAC = angle CAD ( by CPCT

also angle BCA = DCA ( by CPCT

Hence proved AC bisect angle A and C

2) in triangle ABE and ADE

AB = AD ( given

AE = AE( common side

angle A = angle A ( AC bisect angle A

triangle ABE ~ ADE ( by SAS rule

BE = DE ( by CPCT

3) in triangle ABC and ADC

angleABC=ADC ( by CPCT

Answer 2

Answer:

[tex](i) 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 = ∠ ADC.

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