Answer:
The statement is true
Explanation:
we know that
In a Rhombus the diagonals bisect opposite angles.
therefore
diagonal AD bisect angle D and angle A
so
m∠CDA≅m∠BDA
m∠CAD≅m∠BAD
diagonal CB bisect angle C and angle B
m∠ACB≅m∠DCB
m∠ABC≅m∠DBC
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