Question

3
In AABC, DE ll BC and CD ll EF Prove
that AD² = A FX AB
1​

Answer 1

Given :

• In ∆ABC , DE || EF and CD || EF .

To Prove :

• AD² = AF×AB

Theorem to be used :

Basic Proportionality Theorem :

If PQR is a triangle and S and T are points on PQ and PR respectively such that ST || QR then :

$\sf{ \dfrac{PS}{QS} } = \dfrac{PT}{RT}$

Solution :

Applying the theorem in ∆ADC

$\sf \implies \dfrac{DF}{AF} = \dfrac{CE}{AE}\\ \\ \sf \implies \dfrac{DF}{AF} + 1 = \dfrac{CE}{AE} + 1 \\ \\ \implies \sf{\dfrac{AD}{AF} } = \dfrac{AC}{AE} \: \: \: ..........(1)$

And Applying in ∆ABC

$\sf{ \dfrac{BD}{AD} } = \dfrac{CE}{AE} \\ \\ \implies \sf\dfrac{BD}{AD} + 1= \dfrac{CE}{AE} + 1 \\ \\ \implies \sf\dfrac{AB}{AD} = \dfrac{AC}{AE} \: \: ..............(2)$

Now comparing (1) and (2) we have :

$\sf{\dfrac{AD}{AF} = \dfrac{AB}{AD}} \\ \\ \implies \sf{AD \times AD = AF \times AB } \\ \\ \implies \sf {AD}^{2} = AF \times AB$

$\bold{Hence \: \: Proved}$