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ABCD is a trapezium in which AB || CD and AD = BC. Show that : $\left(i\right)\angle A=\angle B\ \left(ii\right)\ \angle C\ =\angle D\ \left(iii\right)\ \Delta ABC\ \cong\Delta BAD$ (iv) diagonal AC = diagonal BD.

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A triangle ABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that ∠BAL=∠ACB

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In fig M and N are two plane mirrors perpendicular to each other; prove that the incident ray CA is parallel to reflected ray BD.

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

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In the given figure, if PQ ∥ PS, ∠SQR = 28° and ∠QRT = 65°,then find the values of x and y

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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 .

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

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

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

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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

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In the given figure, if PQ ∥ ST, ∠PQR= 110

^{0 }and^{ }∠RST= 130^{0}find ∠QRS - Open EndedPlease save your changes before editing any questions.
The polynomial p(x)=x

^{4}−2x^{3}+3x^{2}−ax+3a−7 when divided by x + 1 leave 19 as remainder. Also, find the remainder when p(x) is divided by x + 2. - Open EndedPlease save your changes before editing any questions.
Factorise : $\frac{1}{27}\ \left(2x+5y\right)^3\ +\ \left(\frac{-5}{3}y+\frac{3}{4}z\right)^3-\left(\frac{3}{4}z+\frac{2}{3}x\right)^3$

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