Question

state and prove angle sum property of a quadrilateral

Answer 1

(draw a quad. ABCD )(let ∠DAB=∠1, ∠ADB=∠2, ∠BDC=∠3, ∠DCB=∠4, ∠CBD=∠5, ∠DBA=∠6)

Given: ABCD is a quad.

to prove: ∠A+∠B+∠C+∠D=180°

Construction:Join BD

Proof: In ΔABD,

∠1+∠2+∠6=180° [Angle sum property of triangle]   ........(i)

In ΔBCD,

∠3+∠4+∠5=180° [Angle sum property of triangle]   ..........(ii)

Adding (i) and (ii) we get

∠1+∠2+∠3+∠4+∠5+∠6=180°+180°=360°

∴ ∠A+∠B+∠C+∠D=180°       {∵∠5+∠6=∠B; ∠2+∠3=∠D}

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