Question

In figure, ABCD is a sqaure. M is the midpoint of AB and PQ perpendicular to CM meets AD at P and CB introduced at Q. Prove that

a) ΔPAM congruent to ΔQBM

b) CP = CQ

URGENT!!!

Answer 1

We will first show that Δ PAM is similar to  Δ QBM.

Angle A = 90    and  angle B = 90  as they are angles in a square
angle PMA = angle QMB  as these are the included angles between two straight lines intersecting at one point.

Since two angles in ΔPAM are equal to two corresponding angles in ΔQBM, the third angle in each are also equal.  Hence the two triangles are similar.

Now AM = MB  as M is the midpoint.     So in similar triangles, if a side is equal to a side in the other, they become CONGRUENT.  It means other sides in ΔPAM are also equal to sides in ΔQMB

=============================================
we show that  ΔMPC and ΔMQC are congruent.

Since ΔPMA and ΔQMB are congruent,  side PM = MQ
Side MC is present in both ΔMPC and  ΔMQC.
Angle CMP = angle CMQ  = 90 deg    as  PQ is perpendicular to CM - given

So two sides and one angle in ΔMPC are equal to corresponding sides and angle in  ΔMQC.

So they are congruent.  Hence  CP = CQ.

Answer 2

Answer:

Step-by-step explanation:

given

ABCD is a square

pq perpendicular cm

to prove

in triangle AMP and triangle BMQ

angle PAM = angle QBM

AM = BM

angle PMA = angle QMB

HENCE triangle AMP congruent to triangle BMQ by ASA.

MP = MQ

In triangle CMP and triangle CMQ

CM = CM (common)

angle MP = MQ (proved)

hence triangle CMP congruent to triangle CMQ ny SAS.

Therefore CP = CQ by C.P.C.T.

(HENCE PROVED)