Answer :
Given: ∠1 = ∠2
ΔNSQ≅ΔMTR
To prove: ΔPTS~ΔPRQ
Proof:
Now it is given ΔNSQ≅ΔMTR
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 + 2∠1 = 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 + 2∠PQR = 180° …………(iii)
Equating equation (ii) and (iii), we get
∠P + 2∠1 = ∠P + 2∠PQR
Now cancelling the like terms, we get
2∠1 = 2∠PQR
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
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