in the given figure, AB=CD and AD=BC. PROVE that triangle BAC =ACD.
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given that that AB is equal to CD So AD is parallel to BC so, the angle BAC and angle ACD are the corresponding angles so they are always equal.
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In ΔABC and ΔCDA:
AC=AC (common)
AB=CD (Given)
AD=BC(Given)
By SSS property:
Δ ABC ≅Δ CDA
Thus, ∠BAC =∠ ACD
Hence, proved.
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