Given:$AB \parallel CD, CD\parallel EF$ and $y:z=3:7$.To do:We have to find $x$.Solution:Given, $AB \parallel CD$ and $CD\parallel EF$ We know that, When the angles are on the same side of the transversal line the sum of the angles is always $180^o$.This implies, $x+y=180^o$We also know that,  The corresponding interior angles of ... Read More

Given :$AB \parallel CD$ and $EF$ is perpendicular to $CD$.$\angle GED = 120^o$.To find :We have to find  $\angle GEC, \angle EGF, \angle GEF$ Solution :$\angle GEF + \angle CEG = 120^o$$120^o = \angle GEF + 90^o$$\angle GEF = 120^o-90^o$$\angle GEF = 30^o$$CD$ is a straight line.Therefore, $\angle GED + \angle ... Read More

Given:It is given that $\angle XYZ=64^o$, $XY$ is produced to point $P$ and ray $YQ$ bisects $\angle ZYP$.To do:We have to draw a figure from the given information and find $\angle XYQ$ and reflex $\angle QYP$.Solution:$XYP$ is a line.Therefore, $\angle XYZ+\angle ZYP=180^o$$64^o+\angle ZYP=180^o$  (since $\angle XYZ=64^o$) This implies, $\angle ZYP=180^o-64^o$$\angle ZYP=116^o$Since, ... Read More

To do:We have to find the values of $x$ and $Y$ and then show that $AB \parallel CD$.Solution:We know that, The sum of the measures of the angles in linear pairs is always $180^o$.Since $x$ and $50^o$ are linear pairs of $AB$.We get, $x+50^o=180^o$This implies, $x=180^o-50^o$Therefore, $x=130^o$We also know that, ... Read More

Given:$POQ$ is a line, Ray $OR$ is perpendicular to line $PQ$ and $OS$ is another ray lying between rays $OP$ and $OR$.To do:We have to prove that $\angle ROS = \frac{1}{2}(\angle QOS - \angle POS)$.Solution:Ray $OR \perp POQ$.This implies, $\angle POR = 90^o$$\angle POS + \angle ROS = 90^o$.....…(i)$\angle ROS ... Read More

To do:We have to prove that $AOB$ is a line.Solution:We know that, The sum of the measures of the angles in linear pairs is always $180^o$.So in order to prove that $AOB$ is a straight line, we have to prove that $x+y$ is a linear pair of $AOB$.This implies, $x+y=180^o$We ... Read More

Given:$\angle PQR=\angle PRQ$.To do:We have to prove that $\angle PQS=\angle PRT$.Solution:$SQRT$ is a line.We know that, The sum of the measures of the angles in linear pairs is always $180^o$.$\angle PQS+\angle PQR=180^o$   (as they are linear pairs)$\angle PRT+\angle PRQ=180^o$   (as they are linear pairs)Therefore, $\angle PQR=180^o-\angle PQS$..(i)$\angle PRQ=180^o-\angle PRT$....(ii)Since,  $\angle ... Read More

To do:We have to give a reason to prove that Axiom $5$ is universally true.Solution: For Example:Let us consider a watermelon.It weighs $1\ kg$ when it is whole or complete.Now, let us take a part of it and eat it.Then, if we weigh the part which is left it will be ... Read More

To do:We have to rewrite Euclid's fifth Postulate.Solution:Euclid’s fifth postulate: If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which ... Read More

To do:We have to explain that does Euclid's fifth postulate imply the existence of parallel lines.Solution:Yes, Euclid’s fifth postulate does imply the existence of the parallel lines.When the sum of the interior angles is equal to the sum of the right angles, then the two lines will not meet each ... Read More