Q. 19

# In figure <span l

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 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 above equation, we get

P + PQR + PRQ = 180°

P + 2PQR = 180° …………(iii)

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

P + 21 = P + 2PQR

Now cancelling 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 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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses