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 the 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 the 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 canceling 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:** AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

**Explanation:**

From the above figure, AE is perpendicular to BC,

Consider right – angled ΔABE,

Applying the Pythagoras theorem, we get

AB^{2} = AE^{2} + BE^{2}………. (i)

Consider right – angled ΔAEC,

Applying the Pythagoras theorem, we get

AC^{2} = AE^{2} + EC^{2}………. (ii)

Consider right – angled ΔAED,

Applying the Pythagoras theorem, we get

AD^{2} = AE^{2} + ED^{2}………. (iii)

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

AB^{2} + AC^{2} = AE^{2} + AE^{2} + EC^{2} + BE^{2}

AB^{2} + AC^{2} = 2AE^{2} + EC^{2} + BE^{2}……….(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

AB^{2} + AC^{2} = 2AE^{2} + (BD + ED)^{2} + ( BD – ED)^{2}

Applying the expansion of (a + b)^{2} = a^{2} + b^{2} + 2ab and (a – b)^{2} = a^{2} + b^{2} – 2ab, we get

AB^{2} + AC^{2} = 2AE^{2} + BD^{2} + ED^{2} + 2BD.ED + BD^{2} + ED^{2} – 2BD.ED

⇒ AB^{2} + AC^{2} = 2AE^{2} + 2ED^{2} + 2BD^{2}

⇒ AB^{2} + AC^{2} = 2(AE^{2} + ED^{2} + BD^{2})

Now substituting value from equation (iii), we get

⇒ AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

Hence proved

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