Question

0" id="TexFormula1" title=" \bigstar \: \underline{ \underline{ \tt{Question}}} : " alt=" \bigstar \: \underline{ \underline{ \tt{Question}}} : " align="absmiddle" class="latex-formula"> In the given figure , M and N are the centres of two intersecting circles. Prove that :
i. PQ $\perp$ MN
ii. PR = RQ

~Thanks in advance! <3

Answer 1

Answer:

Step-by-step explanation:

First step is to prove triangle MPN ad MQN are congruent.

MP=MQ, radius of circle M

NP=NQ, radius of circle N

we have two S in SAS  and for A, we need to prove angle MPN = MQN

consider triangle QMP, it is isosceles so angle MPQ = MQP

consider triangle QNP; it is isosceles so angle  QPN = PQN

combining the two above

angle MPN = MPQ+QPN = MQP+PQN = MQN

By SAS, triangle MPN and MQN are congruent

By symmetry, MN is perpendicular to PQ

It also disects PQ so PR = RQ

Answer 2

Answer:

See Below.

Step-by-step explanation:

We are given two intersecting circles with centers M and N.

And we want to prove that: I) PQ ⊥ MN and that II) PR = RQ.

Since MP and MQ are radii of the same circle:

$MP\cong MQ$

Likewise, since NP and NQ are radii of the same circle:

$NP\cong NQ$

And by the Reflexive Property:

$PQ\cong PQ$

Therefore, by SSS Congruence:

$\Delta MPN\cong \Delta MQN$

By CPCTC:

$\displaystyle \angle PMN\cong \angle QMN$

And by the Reflexive Property:

$MR\cong MR$

And since they are the radii of the same circle:

$MP\cong MQ$

Therefore, by SAS Congruence:

$\Delta MPR\cong \Delta MQR$

Therefore, by CPCTC:

$PR\cong RQ$

Note that PQ is a chord in Circle M.

Therefore:

$\text{Segment MN bisects chord PQ }$

In a circle, a segment that passes through the center of the circle that is perpendicular to a chord also bisects the chord.

And conversely, a segment that passes through the center of a circle that bisects a chord in the circle is also perpendicular to the chord.

So:

$\displaystyle PQ\perp MN$