Math, asked by harshu6995, 2 months ago

In the adjoining figure, AB = DC and triangle ABC = triangle DCB.
Prove that :
(i) triangle BAC = CDB
(ii)AE = DE.
Therefore triangle ABC = triangle DBC =》BE = CE.

Please explain. ​

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Answers

Answered by sanikapandya8
5

ABC such that ∠DBC = ∠DCB.

To Prove: AD bisects ∠BAC of ∆ABC.

Proof: In ∆DBC,

∵ ∠DBC = ∠DCB ...(1) | Given

∴ DB = DC ...(2)

| Sides opposite to equal angles of a triangle are equal

In ∆ABC,

∵ AB = AC | Given

∴ ∠ABC = ∠ACB

∴ ∠ABC = ∠ACB ...(3)

| Angles opposite to equal sides of a triangle are equal

Subtracting (1) from (3), we get,

∠ABC - ∠DBC = ∠ACB - ∠DCB

⇒ ∠ABD = ∠ACD ...(4)

In ∆ADB and ∆ADC,

AB = AC | Given

DB = DC | Proved in (2)

∠ABD = ∠ACD | Proved in (4)

∴ ∆ADB ≅ ∆ADC

| SAS congruence rule

∴ ∠DAB = ∠DAC | CPCT

⇒ AD bisects ∠BAC of ∆ABC.

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