in the adjoining finger ,ABC is an isosceles triangle in which AB = AC . if AB and AC are produced to D and E re spectively such that BD = CE, prove that BE = CD
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AS GIVEN THAT ,
AB=AC
∴ANGLE ABC= ANGLE ACB
NOW
Angle ABC+Angle DBC=180 degree {Linear pair}
⇒∠DBC =180-Angle ABC
∴ ∠DBC= ∠ECB=180-∠ABC {∵ ∠ABC=∠ACB} (R)
NOW
BD=CE {given}
BC=CB {common}
∠DBC=∠ECB { from (R)}
∴ ΔBCE CONGURRENCE TO ΔCBD
∴ BE=CD { BY C.P.C.T} prroved.
IF IT IS HELPFUL THEN PLEASE BRAINLIEST THE ANSWER.
AB=AC
∴ANGLE ABC= ANGLE ACB
NOW
Angle ABC+Angle DBC=180 degree {Linear pair}
⇒∠DBC =180-Angle ABC
∴ ∠DBC= ∠ECB=180-∠ABC {∵ ∠ABC=∠ACB} (R)
NOW
BD=CE {given}
BC=CB {common}
∠DBC=∠ECB { from (R)}
∴ ΔBCE CONGURRENCE TO ΔCBD
∴ BE=CD { BY C.P.C.T} prroved.
IF IT IS HELPFUL THEN PLEASE BRAINLIEST THE ANSWER.
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