If two parallel lines are intersected by a transversal, then prove that the bisectors of the interior angles form a rectangle.

In △ABC in given figure, the sides AB and AC of △ABC are produced to points E and D respectively. If bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O, then prove that ∠BOC = 90°   $-\frac{1}{2}$   ∠ A . 

If two lines intersect, prove that the vertically opposite angles are equal

In the given figure, ∠Q>∠R, PA is the bisector of ∠QPR and PM⊥QR. Prove that ∠APM =  $\frac{1}{2}$    (∠Q−∠R)

In given figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA respectively. Find ∠APB.

In fig the side AB and AC of △ ABC are produced to point E And D respectively. If bisector BO and CO of ∠CBE And ∠BCD respectively meet at point O, then prove that ∠ BOC =  $90^0-\frac{1}{2}$ ∠BAC

In the given figure, if PQ ∥ ST, ∠PQR= 110⁰ and ∠RST= 130⁰ find ∠QRS

The polynomial p(x)=x⁴−2x³+3x²−ax+3a−7 when divided by x + 1 leave 19 as remainder. Also, find the remainder when p(x) is divided by x + 2.

Factorise : $\frac{1}{27}\ \left(2x+5y\right)^3\ +\ \left(\frac{-5}{3}y+\frac{3}{4}z\right)^3-\left(\frac{3}{4}z+\frac{2}{3}x\right)^3$  

If x - 3 and  $x-\frac{1}{3}$  are both factors of  $px^2+5x+r,$  then show that p = r