In triangle ABC, AD is perpendicular to BC, AD^2=BD×CD.
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Answer:
Given: △ABC, AD
2
=BD×DC, AD⊥BC
Using Pythagoras Theorem
In Δ ABD
AB
2
=AD
2
+BD
2
AB
2
=BD×DC+BD
2
[GivenAD
2
=BD×DC]
AB
2
=BD×(DC+BD)
AB
2
=BD×BC .......... (i)
In Δ ADC,
AC
2
=AD
2
+DC
2
AC
2
=BD×DC+DC
2
=DC×(BD+DC)
AC
2
=DC×BC ..........(ii)
∴AB
2
+AC
2
=BD×BC+DC×BC [Adding (i) and (ii)]
AB
2
+AC
2
=BC×(BD+DC)
AB
2
+AC
2
=BC×BC=BC
2
⇒∠BAC=90
∘
[Using converse of Pythagoras Theorem]
Step-by-step explanation:
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