Question

2. PQR is a triangle right angled at P and M is a
point on QR such that PM IQR. Show that
PM=QM. MR.​

Answer 1

Answer:

Step-by-step explanation:

ANSWER

In △PMR,

By Pythagoras theorem,

(PR)2  =(PM)  2 +(RM)  2     .......(1)

In △PMQ,

By Pythagoras theorem,

(PQ)2  =(PM)2 +(MQ)2   .......(2)

In △PQR,

By Pythagoras theorem,

(RQ) 2 =(RP)  2 +(PQ)  2   ........(3)

∴ (RM+MQ) 2 =(RP)  2 +(PQ)  2

 ∴ (RM)2 +(MQ)2 +2RM.MQ=(RP)2 +(PQ)  2       ....(4)

Adding 1) and 2) we get,

(PR)  2 +(PQ)2  =2(PM)2 +(RM)2 +(MQ)2           ...(5)

From 4) and 5) we get,

2RM.MQ=2(PM)2

∴ (PM) 2 =RM.MQ

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