Question

a transversal cut two Parallel Lines at a and b the two interior angle at a bisected and so are two interior angle at b the four bisectors from a quadrilateral ACBD prove that. 1= ACBD is a triangle. 2= CD is parallel to original parallel lines​

Answer 1

$\huge{\underline{Solution}}$

To prove → ABCD is a rectangle

AD, CD, AB, BC are bisectors of interior angles formed by transversal line with ∥ line.

$\sf{∠BCA=∠CAB}$

Hence, $\sf{CB∥AB}$

Similarly, $\sf{AB∥CB(∠CAB=∠ACB)}$

(Alternate angles)

Therefore quadrilateral ABCD is a ∥gram as both the pairs of opposite sides are ∥

$\sf{∠b+∠b+∠a+∠a=180°}$

⇒$\sf{2(∠b+∠a)=180°}$

⇒$\sf{∠a+∠b=90°}$

That is ABCD is ∥gram & one of the angle is ⊥ angle.

So, ABCD is a Rectangle.

Answer 2

$\huge{\underline{\orange{Solution}}}$

To prove → ABCD is a rectangle

AD, CD, AB, BC are bisectors of interior angles formed by transversal line with ∥ line.

∠BCA=∠CAB

Hence, CB ∥ AB

Similarly, AB ∥ CB

(Alternate angles)

Therefore quadrilateral ABCD is a ∥gram as both the pairs of opposite sides are ∥

→ ∠b+∠b+∠a+∠a=180°

⇒2(∠b+∠a)=180°

⇒∠a+∠b=90°

That is ABCD is ∥gram & one of the angle is ⊥ angle.

So, ABCD is a Rectangle.