Question

Q.1. In the figure given below,
$angle$
BAC-90° and AD
$is \: perpendicular \: to$
BC. Then,


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Answer 1

Step-by-step explanation:

In △ABD,

AB

2

=BD

2

+AD

2

(Pythagoras theorem)

In △ACD,

AC

2

=CD

2

+AD

2

(Pythagoras theorem)

Add both the equations,

AB

2

+AC

2

=BD

2

+AD

2

+CD

2

+AD

2

BC

2

=BD

2

+AD

2

+2BD×CD−2BD×CD+2AD

2

(Apply pythagoras theorem to triangle ABC)

BC

2

=(BD+DC)

2

−2BD×CD+2AD

2

BC

2

=BC

2

−2BD×CD+2AD

2

AD

2

=BD×CD

Answer 2

Answer:

The question is incomplete..

In the figure given below,∠BAC=90 and AD⊥BC, then AD^2 = BD×DC. Prove that the above given statement is true.

Is this the question you meant..?

Step-by-step explanation:

In △ABD,

AB^2 = BD^2 + AD^2 (Pythagoras theorem)

In △ACD,

AC^2 = CD^2 + AD^2 (Pythagoras theorem)

Add both the equations,

AB^2 + AC^2 = BD^2 + AD^2 + CD^2 + AD^2 + BC^2 = BD^2 + AD^2 + 2BD × CD − 2BD × CD + 2AD^2 (Apply pythagoras theorem to triangle ABC)

BC^2 = (BD + DC)^2 − 2BD × CD + 2AD^2

BC^2 = BC^2 − 2BD × CD + 2AD^2

AD^2 = BD × CD

Hence Proved

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