Given:$AC=AE, AB=AD$ and $\angle BAD=\angle EAC$.To do:We have to show that $BC=DE$.Solution:Let us add $\angle DAC$ on both sides of $\angle BAD=\angle EAC$ we get, $\angle BAD+\angle DAC=\angle EAC+\angle DAC$This implies, $\angle BAC=\angle EAD$We know that, According to Rule of Side-Angle-Side Congruence:Triangles are said to be congruent if any pair ... Read More

Given:$AB$ is a line segment and $P$ is its mid-point.$D$ and $E$ are points on the same side of $AB$ such that $\angle BAD=\angle ABE$ and $\angle EPA=\angle DPB$.To do:We have to show that (i) $\triangle DAP \cong \triangle EBP$(ii) $AD=BE$.Solution:(i) Let us add $\angle DPE$ on both sides of $\angle ... Read More

Given:Line $l$ is the bisector of an angle $\angle A$ and bisector $B$ is any point on $l$.$BP$ and $BQ$ are perpendiculars from $B$ to the arms of $\angle A$.To do:We have to show that(i) $\triangle APB \cong \triangle AQB$(ii) $BP=BQ$ or $B$ is equidistant from the arms of $\angle ... Read More

Given:$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$.To do:We have to show that $\triangle ABC\cong \triangle CDA$.Solution:Let us consider $\triangle ABC$ and $\triangle CDA$, We know that, When the lines intersected by the transversal are parallel, alternate interior angles are equal.This ... Read More

Given:$AD$ and $BC$ are equal perpendiculars to a line segment $AB$.To do:We have to show that $CD$ bisects $AB$.Solution:We know that, From Angle Angle Side congruence rule: If two pairs of corresponding angles along with the opposite or non-included sides are equal to each other then the two triangles are ... Read More

(a) Air - mixture(b) Copper - pure substance(c) Silver - pure substance(d) A sugar solution - mixture(e) Wind - mixture(f) A salt solution - mixture(g) Carbon dioxide - mixture(h) Water - mixture(i) Nitrogen - pure subs.(j) Iron - pure substance(k) Oxygen - pure substance(l) milk - mixture(m) Blood - mixture(n) ... Read More

Given:In quadrilateral $ABCD, AC=AD$ and bisects $\angle A$.To do:We have to show that $\triangle ABC\cong ABCD$ and say about $BC$ and $BD$.Solution:Let us consider $\triangle ABC$ and $\triangle ABD$.Given, $AC=AD$The line segment $AB$ bisects $\angle A$.Therefore, $\angle CAB=\angle DAB$We know that, According to Rule of Side-Angle-Side Congruence:Triangles are said to ... Read More

Given:The side $QR$ of $\triangle PQR$ is produced to a point $S$.The bisectors of $\angle PQR$ and $\angle PRS$ meet at a point $T$.To do:We have to prove that $\angle QTR=\frac{1}{2}\angle QPR$.Solution:Let us consider the $\angle PQR$ We know that, The sum of the interior angles is equal to the exterior ... Read More

Given:$PQ \perp PS, \angle SQR=28^o$ and $\angle QRT=65^o$.To do:We have to find the values of $x$ and $y$.Solution:Since, $QR$ is a transversal, the alternate angles are added.$x+\angle SQR=\angle QRT$By substituting the values of $\angle QRT$ and $\angle SQR$ we get, $x+28^o=65^o$$x=65^o-28^o$$x=37^o$We also know that, The lines intersected by the transversal ... Read More

Given:$AB \parallel DE, \angle BAC=35^o$ and $\angle CDE=53^o$.To do:We have to find $\angle DCE$.Solution:We know that, The lines $AB \parallel DE$Therefore, $AE$ becomes the transversal of $AB$ and $DE$.Since the lines intersected by the transversal are parallel, alternate interior angles are equal.This implies, $\angle BAC=\angle AED$Since the value of $\angle ... Read More