In triangle abc ad is perpendicular to bc and point d lies on bc , bd =3cd then prove that
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Given: AD is the median of ΔABC.
To prove: AB + AC > 2AD
Construction: Produce AD to E such that AD = DE. Join CE.
Proof: In ΔABD and ΔCDE
AD = DE (Construction)
∠ADB = ∠CDE (Vertically opposite angles)
BD = DC (AD is the median from A to BC)
∴ ΔABDΔCDE (SAS congruence criterion)
⇒ AB = CE (CPCT) ...(1)
In ΔACE,
AC + CE > AE (Sum of any two sides of a triangle is greater than the third side)
⇒ AC + AB > AD + DE [Using (1)]
⇒ AC + AB > AD + AD (Constriction)
⇒ AC + AB > 2AD
To prove: AB + AC > 2AD
Construction: Produce AD to E such that AD = DE. Join CE.
Proof: In ΔABD and ΔCDE
AD = DE (Construction)
∠ADB = ∠CDE (Vertically opposite angles)
BD = DC (AD is the median from A to BC)
∴ ΔABDΔCDE (SAS congruence criterion)
⇒ AB = CE (CPCT) ...(1)
In ΔACE,
AC + CE > AE (Sum of any two sides of a triangle is greater than the third side)
⇒ AC + AB > AD + DE [Using (1)]
⇒ AC + AB > AD + AD (Constriction)
⇒ AC + AB > 2AD
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