Question

PLEASE HELP ASAP
In △MNP, point Q is on side MN so that PM=PQ. Prove that PN>PM.

Answer 1

Answer:

Given: In $\triangle MNP$, Point Q is on the side MN so that PM=PQ.

Isosceles Triangle states:

* A triangle with two equal sides.

*The angles which are opposite to the equal sides are also equal.

Since, PQ=PM

By definition of isosceles triangle,

$\triangle PMQ$ is an isosceles triangle,

All the three angles within an isosceles triangle are acute. Acute angles are always less than 90 degrees.

therefore,  $\angle PMQ=\angle PQM$$<90^{\circ}$    ......[1]

Now, Consider $\triangle PNQ$,

Exterior Angle Property state that  an exterior angle of a triangle is equal to the sum of the opposite interior angles.

Therefore, by exterior angle property in $\triangle PNQ$ we have ;

$\angle PNQ + \angle NPQ = \angle PQM$

So, $\angle PNQ< \angle PQM$

From [1] we have:

$\angle PNQ < \angle PQM=\angle PMQ <90^{\circ}$

To prove PN>PM.

Consider  $\triangle PMN$,

Sine law states that an equation relating the lengths of the sides of a triangle to the sines of its angles.

Now, using Sine law in  $\triangle PMN$,

$\frac{\sin (\angle PMQ)}{PN} =\frac{\sin (\angle PNQ)}{PM}$      ......[2]

Since,

$\angle PNQ < \angle PQM< 90^{\circ}$

so,  $\sin (\angle PNQ)<\sin (\angle PMQ)$                         ......[3]

From [2] and [3] we have;

$PM<PN$ or $PN>PM$     hence Proved!