Question

Please solve both parts of Qno. 19

Answer 1

a.

Given, ∠1 = ∠2

According to the properties of scalene triangles, if bases angles are equal then sides containing them become equal.

∴ PT = PS           ...( i )

 

As ΔNSQ ≅ ΔMTR,  

By c.p.c.t ( corresponding part of congruent triangle ),  

SQ = TR           ...( ii )  

In ( i ), divide both sides by SQ.

$\implies \dfrac{PT}{SQ} = \dfrac{PS}{SQ}$

From ( ii ), SQ = TR,  

$\implies \dfrac{PS}{SQ} =\dfrac{PT}{TR}$

Thus in ΔPQR, $\dfrac{PS}{SQ}=\dfrac{PT}{TR}$  

By the converse of Thales theorem,

⇒ ST ║ QR

∴ ∠1 = ∠Q ( Corresponding angles )

In ΔPTS and ΔPRQ,

∠1 = ∠Q

∠SPT = ∠QPR

Two angles are equal, the remaining angles will also be equal.

Hence, proved that ΔPTS $\sim$ ΔPRQ by AA similarity criterion.

b.  

In Δ ΔAMD, ∠AMD = 90°

By Pythagoras theorem,

⇒ AD^2 = AM^2 + MD^2  

⇒ AM^2 = AD^2 - MD^2     ...( i )

In ΔAMC, ∠AMC = 90°

By Pythagoras theorem,

⇒ AC^2 = AM^2 + MC^2

As MC is sum of MD and DC, ( MC )^2 = ( MD + DC )^2. Substitute the value of AM^2 from ( i ).

⇒ AC^2 = AD^2 - MD^2 + ( MD + DC )^2  

⇒ AC^2 = AD^2 - MD^2 + ( MD + BC / 2 )^2

⇒ AC^2 = AD^2 - MD^2 + MD^2 + ( BC / 2 )^2 + MD.BC

⇒ AC^2 = AD^2 + MD.BC + ( BC^2 / 4 )     ...( ii )

In ΔABM, ∠AMB = 90°

By Pythagoras theorem,

⇒ AB^2 = AM^2 + BM^2

⇒ AB^2 = ( AD^2 - MD^2 ) + ( BD - MD )^2

⇒ AB^2 =  AD^2 - MD^2 + ( BC / 2 - MD )^2

⇒ AB^2 = AD^2 - MD^2 + ( BC^2 ) / 4 + MD^2 - BC.MD

⇒ AB^2 = AD^2 - BC.MD + ( BC^2 ) / 4    ...( iii )

Adding ( ii ) & ( iii ),

⇒ AB^2 + AC^2 = AD^2 + MD.BC + ( BC^2 / 4 ) +  AD^2 - BC.MD + ( BC^2 ) / 4

⇒ AB^2 + AC^2 = 2AD^2 + ( BC^2 / 2 )

⇒ AB^2 + AC^2 = 2AD^2 + 1 / 2 ( BC^2 )

⇒ AB^2 + AC^2 = 2AD^2 + 1 / 2 ( 2 BD^2 )

⇒ AB^2 + AC^2 = 2AD^2 + 1 / 2 ( 4 BD^2 )

⇒ AB^2 + AC^2 = 2AD^2 + 2BD^2

⇒ AB^2 + AC^2 = 2( AD^2 + BD^2 )

Hence, proved.