In an equilateral triangle ABC , D is a point on side BC, such that BD =1/3 BC. Prove that 9AD square = 7 AB square.
Solve this by taking LHS or RHS..
Answers
Answer: Given ABC is an equilateral triangle, BD = 1/3BC
Let AB = BC = AC = "a" ----------------------(since equilateral triangle has all sides equal)
CONSTRUCTION: Draw AL perpendicular to BC.
We also know that in an equilateral triangle ,perpendicular drawn from one of it's vertex to another side ,bisects the side.
⇒BL = LC -------------------(1.)
Also given that,
BD = 1/3BC = 1/3a
⇒CD = 2/3 BC =2/3a
Also,
BL = 1/2 BC -----------------------------(by 1.)
BL =1/2a
Also,
DL = BL - BD
⇒DL = a/2 - a/3 = a/6 -----------------------------A.
IN TRIANGLE ALD,
AD² = AL² + LD² (pythagoras theorem) --------------------------B.
IN TRIANGLE ALB,
AB² = a² = AL² + LB² ------------------------C.
⇒AL² = AB² - LB² ---------------------D.
Putting the relations in B. we get,
AD² = a² - a²/4 +a²/36 = 36a² - 9a² + a²/36 (AFTER LCM)
⇒AD² = 28a²/36 = 7a²/9
⇒9AD² = 7a²
Or,
9AD² = 7AB² (SINCE , a² = AB² = AC² = BC²)
HENCE PROVED!!!!!!
HOPE THIS HELPS ...
Answer:construction ,(i) draw AE ⊥BC (ii) join AD
IN ΔAEB and AEC,
AB=AC ( ABC IS EQUILATERAL Δ)
∠AEB=∠AEC ( EACH EQUAL TO 90° )
AE=AE ( COMMON )
Δ AEB≅ ΔAEC (RHS)
⇒ BE=EC (CPCT)
NOW,
BD=1/3BC , DC=2/3BC AND BE=EC=1/2BC
SINCE ∠C=60°. Therefore ADC is acute triangle
⇒ AD²=AC²+DC²-2DC×EC
AD²=AC²+2/3BC²-2×2/3BC×1/2BC ( DC=2/3BC , EC=1/2BC )
AD²=AC²+4/9 BC²-2/3BC²
AD²=AB²+4/9AB²-2/3AB² ( AB=AC=BC )
AD²=9AB²+4AB²-6AB²/9=7/9 AB²
AD²=7/9 AB²
9AD²=7 AB²
IN ΔAEB and AEC,
AB=AC ( ABC IS EQUILATERAL Δ)
∠AEB=∠AEC ( EACH EQUAL TO 90° )
AE=AE ( COMMON )
Δ AEB≅ ΔAEC (RHS)
⇒ BE=EC (CPCT)
NOW,
BD=1/3BC , DC=2/3BC AND BE=EC=1/2BC
SINCE
Step-by-step explanation: