Question

Chapter 6: Triangles
1. In the fig, DE ll FC, and FE II BC prove that AF^2=AD x AB

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Answer 1

Given

DE || FC and FE || BC

To Prove

AF² = AD × AB

Proof

In ∆AFC,

AD/DF = AE/EC (Basic Proportionality Theorem)

→ DF/AD = EC/AE

Adding 1 on both sides,

(DF/AD) + 1 = (EC/AE) + 1

→ (AD + DF)/AD = (AE + EC)/AE

→ AF/AD = AC/AE ...........(eq 1)

Now, in ∆ABC,

AF/FB = AE/EC (Basic Proportionality Theorem)

→ FB/AF = EC/AE

Add 1 on both sides

→ (FB/AF) + 1 = (EC/AE) + 1

→ (AF + FB)/AF = (AE + EC)/AE

→ AB/AF = AC/AE ............(eq 2)

From (eq 1) and (eq 2),

AF/AD = AB/AF

Cross multiplying, we get

AF² = AB × AD

PROVED

Answer 2

Given: ∆ABC having DE || FC and FE || BC

Prove: AF² = AD × AB

Proof:

In ∆ABC

FE || BC

So,

→ $\dfrac{FB}{AF}\:=\:\dfrac{EC}{AE}$ (By BPT)

{BPT - Basic Proportionality Theoram}

Now.. Add 1 on both sides

→ $\dfrac{FB}{AF}\:+\:1\:=\:\dfrac{EC}{AE}\:+\:1$

→ $\dfrac{FB\:+\:AF}{AF}\:=\:\dfrac{EC\:+\:AE}{AE}$

Now.. FB + AF = AB and EC + AE = AC

→ $\dfrac{AB}{AF}\:=\:\dfrac{AC}{AE}$ ________ (eq 1)

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Now; In ∆AFC

DE || FC

So,

→ $\dfrac{AD}{DF}\:=\:\dfrac{AE}{EC}$ (By BPT)

Do reciprocal of it

→ $\dfrac{DF}{AD}\:=\:\dfrac{EC}{AE}$

Add 1 on both sides

→ $\dfrac{DF}{AD}\:+\:1\:=\:\dfrac{EC}{AE}\:+\:1$

→ $\dfrac{AD\:+\:DF}{AD}\:=\:\dfrac{AE\:+\:EC}{AE}$

→ $\dfrac{AF}{AD}\:=\:\dfrac{AC}{AE}$ _____ (eq 2)

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From (eq 1) and (eq 2) we get;

→ $\dfrac{AB}{AF}\:=\:\dfrac{AF}{AD}$

Cross-multiply them

→ (AB)(AD) = (AF)(AF)

→ AB × AD = AF²

→ AF² = AB × AD

→ AF² = AD × AB

_________ [ PROVED ]

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