Question

ABCD is a rhombus. Show that diagonal AC bisects Z A as well as 2 C and diagonal BD bisects < B as well as D. i need step by step explanation
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Answer 1

Answer:

In △ABD and △BCD,

AB=BC ....Sides of a rhombus

AD=CD ....Sides of a rhombus

AC=AC ....Common side

△ABD≅△BCD ...SSS test of congruence

∴∠ABD=∠CBD ....C.P.C.T.

∴∠ADB=∠CDB ....C.P.C.T.

So, Diagonal BD bisects ∠B and ∠D of the rhombus ABCD.

Similarly we can prove,

△ABC≅△ADC

∴∠BAC=∠DAC ....C.P.C.T.

∴∠BCA=∠DCA ....C.P.C.T

So, Diagonal AC bisects ∠A and ∠C of the rhombus ABCD.