Question

ABCD is a square. M is the point on AB and PQ  CM. Prove that CP = CQ.

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

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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.

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)

Answer 2

Answer:

CP = CG both are equal only