Question

In figure 13 , AC , AD and AE are joined . Prove that angle FAB + angle ABC + angle BCD + angle CDE +angle DEF + angle EFA = 720 degree.​

Answer 1

Answer:

in any quadrilateral or AFED sum of all angles is always 360 .and in quadrilateral ABCD also sum of all angles is always 360 hence....sum of both the quadrilaterals = 360+360 = 720

...joining of lines and all are just given for confusion

Step-by-step explanation:

hope it' help

Answer 2

Answer with Step-by-step explanation:

AD and AE are joined.

In triangle AFE

$\angle AFE+\angle FAE+\angle AEF=180^{\circ}$..(1)

By using triangle angles sum property

In triangle AED

$\angle AED+\angle DEA+\angle ADE=180^{\circ}$...(2)

In triangle ADC

$\angle ACD+\angle ADC+\angle CAD=180^{\circ}$..(3)

In triangle ABC

$\angle ABC+\angle CAB+\angle ACB=180^{\circ}$

Adding equation (1),(2),(3) and (4)$\angle AFE+\angle FAE+\angle FEA+\angle AED+\angle EDA+\angle DAE+\angle ADC+\angle DCA+\angle CAD+\angle ABC+\angle ACB+\angle CAB=720^{\circ}$$\angle AFE+(\angle FAE+\angle DAE+\angle CAD+\angle CAB)+(\angle FEA+\angle AED)+(\angle EDA+\angle ADC)+(\angle ACB+\angle ACD)+\angle ABC=720^{\circ}$$\angle FAB+\angle ABC+\angle BCD+\angle CDE$+$\angle$DEF+$\angle EFA$=$720^{\circ}$

Hence, proved.

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