Question

in the fig ac =bd and ac parallel to db .prove that triangle abc=triangle bpd

Answer 1

Solution:-
We have,
∠ AOC = ∠ ACO and ∠ BOD = ∠ BDO
But, ∠ AOC ∠ BOD (Vertically opposite angles)
∴ ∠ ACO = ∠ BOD and ∠ BOD = ∠ BDO
⇒ ∠ ACO = ∠ BDO
Thus AC and BD are two lines intersected by transversal CD such that       ∠ ACO = ∠ BDO i.e. alternate angles are equal. Therefore AC II DB

Answer 2

$\underline{\underline{\Huge\mathfrak{Answer ;}}}$

Dear ,

We have ;

∠ AOC = ∠ ACO and ∠ BOD = ∠ BDO
But, ∠ AOC ∠ BOD (Vertically opposite angles)

∴ ∠ ACO = ∠ BOD and ∠ BOD = ∠ BDO
⇒ ∠ ACO = ∠ BDO


• Thus AC and BD are two lines intersected by transversal CD such that  ∠ ACO = ∠ BDO i.e. alternate angles are equal.

So , AC II DB { Proved }

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@dmohit432