# Plzz solve 3,4,5,6...URGENT

Solutions for the questions are as under:

**Answer (3):**In the given figure, we can see that;

$\angle BAD=\angle EAC(giveninthequestion)\phantom{\rule{0ex}{0ex}}adding\angle DACtoboththesides,weget,\phantom{\rule{0ex}{0ex}}\angle BAD+\angle DAC=\angle EAC+\angle DAC\phantom{\rule{0ex}{0ex}}or,\phantom{\rule{0ex}{0ex}}\angle BAC=\angle EAD.....\left(i\right)(as\angle BAD+\angle DAC=\angle BACand\angle EAC+\angle DAC=\angle EAD)$

Hence, considering $\u25b3ABCand\u25b3ADE$

$\angle BAC=\angle EAD\left\{From\left(i\right)\right\}\phantom{\rule{0ex}{0ex}}AC=AE\left\{Giveninthequestion\right\}\phantom{\rule{0ex}{0ex}}AB=AD\left\{Giveninthequestion\right\}$

Thus, based on SAS (Side Angle Side) property we can say that $\u25b3ABCand\u25b3ADE$ are congruent with each other.

Therefore, BC = DE (the third side of the congruent triangles)

**Answer (4):**It is given in the question that $\u25b3ABCand\u25b3RPQ$ are congruent with each other. Thus, the corresponding sides which are equal with each other as per the properties of congruency, will be;

$AB=RP\phantom{\rule{0ex}{0ex}}BC=PQ\phantom{\rule{0ex}{0ex}}AC=QR$

But $BC$ will not be equal to $QR$, as these are not the corresponding sides. Thus it is false to say that $BC=QR$.

**Answer (5):**It is given in the question that $\u25b3PQRand\u25b3EDF$ are congruent with each other. Thus, the corresponding sides which are equal with each other as per the properties of congruency, will be;

$PQ=ED\phantom{\rule{0ex}{0ex}}QR=DF\phantom{\rule{0ex}{0ex}}PR=EF$

Hence, we can definitely say that $PR$ will be equal to $EF$, as these are the corresponding sides. Thus it is true to say that $PR=EF$.

**Answer (6):**It is given in the question that angles A, B and C of the triangle are equal with each other. Hence we can also say that all the three angles are congruent with each other.

Thus,

$\angle A\cong \angle B\cong \angle C$

Now, in a triangle, sides opposite to the equal and congruent angles are always equal with each other.

Hence,

$BC=AC(Sidescorrespoingtoequalangles\angle Aand\angle B)\phantom{\rule{0ex}{0ex}}AC=AB(Sidescorrespoingtoequalangles\angle Band\angle C)\phantom{\rule{0ex}{0ex}}BC=AB\left(\right(Sidescorrespoingtoequalangles\angle Aand\angle C)$

Therefore, we can that $AB=BC=AC$.

Since, in $\u25b3ABC$, all the angles are equal and all the sides are equal, thus $\u25b3ABC$ is an equilateral triangle.

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards

**
**