Given : In ∆ABC,AD⊥BC and AD2=BD×DC To find : ∠BAC Solution : In right ∆ADB and ∆ADC, if we apply Pythagoras Theorem, ⇒ (AB)2=(AD)2+(BD)2 ---(i) and ⇒ (AC)2=(AD)2+(DC)2 ------(ii) Adding equations (i) and (ii), we get : ⇒ (AB)2+(AC)2=2(AD)2+(BD)2+(DC)2 ⇒ (AB)2+(AC)2=2(BD)(DC)+(BD)2+(DC)2 [Given] ⇒ (AB)2+(AC)2=(BD+DC)2 ⇒ (AB)2+(AC)2=(BC)2 Hence, ∆ABC is a right triangle right angled at A. ∴ ∠BAC = 90°