Question

In the figure , angle BAC =90° and AD perpendicular to BC. Prove that

(a) BD×CD=BC ^2
(b) AB×AC=BC ^2
(c) BD×CD =AD ^2
(d)AB×AC=AD^2

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Answer 1

EXPLANATION.

< BAC = 90°   ( given ).

AD ⊥ BC.  ( given ).

Pythagoras theorem,

H² = B² + P².

in ΔABD

AB² = BD² + AD²  ......(1).

in ΔACD

AC² = AD² + DC²    .......(2).

BD + DC = BC.

METHOD = 1.

Add equation (1) and (2) we get,

⇒ AB² + AC² = BD² + AD² + CD² + AD².

⇒ AB² + AC² = BD² + AD²  + 2BD X CD - 2BD X CD + 2AD².

⇒ BC² = ( BD + CD )² - 2BD X  CD + 2AD².

⇒ BC² = BC² - 2BD X CD + 2AD².

⇒ AD² = BD X CD.  

METHOD = 2.

⇒ < ADB = < ADC  = [ Each are = 90° ].

⇒ < DBA = < DAC [ by AAA criteria ].

⇒ ΔDBA = ΔDAC.

⇒ ΔDBA ≅ ΔDAC [ AAA criteria ].

⇒ DB/DA = DA/DC.

⇒ BD/AD = AD/DC.

⇒ AD² = BD X DC

HENCE PROVED.