In figure, AC - AE, AB - AD and BAD = EAC. Show that BC - DE
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Ex 7.1, 6 In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE. Given: AC = AE AB = AD ∠ BAD = ∠ EAC ...
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We have, ∠BAD = ∠EAC
Adding ∠DAC on both sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠DAE
Now, in ∆ABC and ∆ADE. we have
∠BAC = ∠DAE [Proved above]
AB = AD [Given]
AC = AE [Given]
∴ ∆ABC ≅ ∆ADE [By SAS congruency]
⇒ BC = DE [By C.P.C.T.]
Step-by-step explanation:
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