20) In A ABC, AB = AC and BD I AC, prove that
BD 2 + CD 2 = 2AC
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Answer:
Answer
Given AB=AC,BD⊥AC
From right △ADB,
By pythagorus theorem, AB
2
=AD
2
+BD
2
⟹AC
2
=AD
2
+BD
2
[∵AB=AC]
⟹(AD+DC)
2
=AD
2
+BD
2
⟹AD
2
+2.AD.DC+DC
2
=AD
2
+BD
2
⟹2.AD.DC=BD
2
−CD
2
Hence, proved
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