Question

ABCD is a rhombus. Show that the diagonal AC bisects A as well as C and diagonal
BD bisects B as well as D.

Answer 1

triangle abc=~to triangle adc

so angles bac=dac

dca=bca

similarly, triangles abd=~cbd

so,angles,abd=cbd

adb=cdb

Answer 2

$\huge\underline\mathbb{ANSWER}$

In trinagle ABC and traingle ADC

=> AB = AD (As ABCD is a rhombus)

=> BC = CD (As ABCD is a rhombus)

=> AC = AC (common)

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By SSS rule we get,

=> trinagle ABC congurent to triangle ADC

=> angle BAC = angle DAC

=> angle ACB = angle ACD

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Diagonal AC bisects angle A as well as angle C (CPCT)

Diagonal BD bisects angle B as well as angle D (CPCT)

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