in the given figure. if AC = AE, AB = AD and BAD = EAC, then
3
D
(1) ADE - ABD
(2) BC - DE
(3) AD AC
(4) Both (1) & (2)
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Step-by-step explanation:
It is given that ∠BAD=∠EAC
∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]
∴∠BAC=∠DAE
In △BAC and △DAE
AB=AD (Given)
∠BAC=∠DAE (Proved above)
AC=AE (Given)
∴△BAC≅△DAE (By SAS congruence rule)
∴BC=DE
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