Q. 193.7( 3 Votes )

In figure <span l

Answer :

Given: 1 = 2


ΔNSQΔMTR


To prove: ΔPTS~ΔPRQ


Proof:


Now it is given ΔNSQΔMTR


So, by CPCT (corresponding parts of congruent triangles are equal), we have


NQS = MRT


Or it can also be written as


PQR = PRQ………….(i)


Now consider ΔPST,


We know in a triangle sum of all the angles is equal to 180°, hence


P + PST + STP = 180°


Or, P + 1 + 2 = 180°


But given 1 = 2, substituting this value in the above equation, we get


P + 1 + 1 = 180°


P + 21 = 180°…………(ii)


Now consider ΔPQR,


We know in a triangle sum of all the angles is equal to 180°, hence


P + PQR + PRQ = 180°


Substituting value from equation (i) in the above equation, we get


P + PQR + PRQ = 180°


P + 2PQR = 180° …………(iii)


Equating equation (ii) and (iii), we get


P + 21 = P + 2PQR


Now canceling the like terms, we get


21 = 2PQR


Or, 1 = PQR…………(iv)


Or, PST = PQR


Now consider ΔPTS and ΔPRQ,


P = P (common)


PST = PQR (from equation (iv))


Hence the AA similarity,


ΔPTS ~ ΔPRQ


Hence proved


OR


Given: in the above figure ΔABC, if AD is the median


To show: AB2 + AC2 = 2(AD2 + BD2)


Explanation:


From the above figure, AE is perpendicular to BC,


Consider right – angled ΔABE,


Applying the Pythagoras theorem, we get


AB2 = AE2 + BE2………. (i)


Consider right – angled ΔAEC,


Applying the Pythagoras theorem, we get


AC2 = AE2 + EC2………. (ii)


Consider right – angled ΔAED,


Applying the Pythagoras theorem, we get


AD2 = AE2 + ED2………. (iii)


Adding equation (i) and (ii), we get


AB2 + AC2 = AE2 + AE2 + EC2 + BE2


AB2 + AC2 = 2AE2 + EC2 + BE2……….(iv)


But from figure


BE = BD – ED and EC = CD + ED


or EC = BD + ED (as AD is the median)


Substituting the above values in equation (iv), we get


AB2 + AC2 = 2AE2 + (BD + ED)2 + ( BD – ED)2


Applying the expansion of (a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 – 2ab, we get


AB2 + AC2 = 2AE2 + BD2 + ED2 + 2BD.ED + BD2 + ED2 – 2BD.ED


AB2 + AC2 = 2AE2 + 2ED2 + 2BD2


AB2 + AC2 = 2(AE2 + ED2 + BD2)


Now substituting value from equation (iii), we get


AB2 + AC2 = 2(AD2 + BD2)


Hence proved

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