Question

7. ABCD is a rhombus. Show that diagonal AC
bisects < A as well as angleC and diagonal BD
bisects < B as well as angleD​

Answer 1

Answer:

[ '<' means angle]

Given : ABCD is a rhombus

To prove: <1 = <2 and <3 = <4

<5 = <6 and <7 = <8

Proof:

In triangle ADC and triangle ABC

AD=AB ( adjacent sides of rhombus are equal)

CD=CB ( adjacent sides of rhombus)

AC=AC ( common)

therefore, triangle ADC =~ triangle ABC by SSS rule

<1 = <2 ( cpct)

<3= <4 ( cpct)

Similarly,

In triangle ABD and triangle CBD

AD= DC ( adjacent sides of rhombus)

AB=BC ( adjacent sides of rhombus)

BD=CD ( common)

therefore, triangle ABD =~ triangle CBD by SSS rule

<5 = <6 ( cpct)

<7= <8 ( cpct)

Hence proved..

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