Question

In given figure angle1 =angle2 and NSQ is congruent to MTR , then prove that PTS ~ PRQ .
OR
In an equilateral triangle ABC, D is a point on the side BC such that BD=1/3BC. Prove that 9AD^2 = 7AB^2

Answer 1

Given: In an equilateral triangle ΔABC. The side BC is trisected at D such that BD = (1/3) BC.

To prove: 9AD2 = 7AB2

Construction: Draw AE ⊥ BC.

Proof :

In a ΔABC and ΔACE

AB = AC ( Given)

AE = AE ( common)

∠AEB = ∠AEC = 90°

∴ ΔABC ≅ ΔACE ( For RHS criterion)

BE = EC (By C.P.C.T)

BE = EC = BC / 2

In a right angled triangle ADE

AD2 = AE2 + DE2 ---------(1)

In a right angled triangle ABE

AB2 = AE2 + BE2 ---------(2)

From equ (1) and (2) we obtain

⇒ AD2 - AB2 = DE2 - BE2 .

⇒ AD2 - AB2 = (BE – BD)2 - BE2 .

⇒ AD2 - AB2 = (BC / 2 – BC/3)2 – (BC/2)2

⇒ AD2 - AB2 = ((3BC – 2BC)/6)2 – (BC/2)2

⇒ AD2 - AB2 = BC2 / 36 – BC2 / 4 ( In a equilateral triangle ΔABC, AB = BC = CA)

⇒ AD2 = AB2 + AB2 / 36 – AB2 / 4

⇒ AD2 = (36AB2 + AB2– 9AB2) / 36

⇒ AD2 = (28AB2) / 36

⇒ AD2 = (7AB2) / 9

9AD2 = 7AB2 .

Answer 2

Step-by-step explanation:

➡ Given :-

→ A ∆ABC in which AB = BC = CA and D is a point on BC such that BD = ⅓BC.

➡ To prove :-

→ 9AD² = 7AB² .

➡ Construction :-

→ Draw AL ⊥ BC .

➡ Proof :-

In right triangles ALB and ALC, we have

AB = AC ( given ) and AL = AL ( common )

∴ ∆ALB ≅ ∆ ALC [ By RHS axiom ] .

So, BL = CL .

Thus, BD = ⅓BC and BL = ½BC .

In ∆ALB, ∠ALB = 90° .

∴ AB² = AL² + BL² .......(1) [ by Pythagoras' theorem ] .

In ∆ALD , ∠ALD = 90° .

∴ AD² = AL² + DL² . [ by Pythagoras' theorem ] .

⇒ AD² = AL² + ( BL - BD )² .

⇒ AD² = AL² + BL² + BD² - 2BL.BD .

⇒ AD² = ( AL² + BL² ) + BD² - 2BL.BD .

⇒ AD² = AB² + BD² - 2BL.BD. [ using (1) ]

⇒ AD² = BC² + ( ⅓BC )² - 2( ½BC ). ⅓BC .

[ ∵ AB = BC, BD = ⅓BC and BL = ½BC ] .

⇒ AD² = BC² + 1/9BC² - ⅓BC² .

⇒ AD² = 7/9BC² .

⇒ AD² = 7/9AB² [ ∵ BC = AB ] .

$\huge \green{ \boxed{ \sf \therefore 9AD^{2} = 7AB^{2}. }}$

Hence, it is proved.