In triangle ABC if AD is the median then show that AB^2+AC^2=2(AD^2+BD^2)
Answers
Given
Triangle ABC with median AD
To prove
+
= 2(
+
)
Construction
Draw a line AM ⊥ BC.
Proof
Case 1
When , it means AD is perpendicular on BC and both angles are right angle i.e, 90°
Then, In ∆ADB,
According to Pythagoras theorem,
AB² = AD² + BD² ..... (1)
In ∆ADC ,
According to Pythagoras theorem,
AC² = AD² + DC² ...... (2)
AD is median.
So, BD = DC .......(3)
From equations (1) , (2) and (3),
AB² + AC² = AD² + AD² + BD² + BD²
AB² + AC² = 2(AD² + BD²)
Hence Proved.
Case 2
Let us consider that, ADB is an obtuse angle.
In ∆ABM,
By Pythagoras theorem,
AB² = AM² + BM²
AB² = AM² + (BD + DM)²
AB² = AM² + BD² + DM² + 2BD.DM ......(1)
In ∆ACM,
According to Pythagoras theorem,
AC² = AM² + CM²
AC² = AM² + (DC - DM)²
AC² = AM² + DC² + DM² - 2DC.DM ......(2)
From equations (1) and (2),
AB² + AC² = 2AM² + BD² + DC² + 2DM² + 2BD.DM - 2DC.DM
AB² + AC² = 2(AM² + DM²) + BD² + DC² + 2(BD.DM - DC.DM) ...........(3)
According to question, AD is median on BC.
So, BD = DC .....(4)
And in ∆ADM,
According to Pythagoras theorem,
AD² = AM² + DM² ........(5)
Putting equation (4) and equation (5) in equation (3),
AB² + AC² = 2AD² + 2BD² + 2(BD.DM - BD.DM)
AB² + AC² = 2(AD² + BD²)
Hence Proved.
Answer:
AB² + AC² = 2(AD² + BD²).
Step-by-step explanation:
(i)
In ΔAED,
⇒ AD² = AE² + DE²
⇒ AE² = AD² - DE²
(ii)
In ΔAEB,
⇒ AB² = AE² + BE²
= AD² - DE² + BE²
= AD² - DE² + (BD + DE)² {BE = BD + DE}
= AD² - DE² + BD² + DE² + 2BD * DE - DE²
= AD² + BD² + 2BD * DE
(iii)
In ΔAEC,
⇒ AC² = AE² + EC²
= AD² - DE² + EC²
= AD² - DE² + (DC - DE)²
= AD² - DE² + DC² + DE² - 2DC * DE
= AD² + BD² - 2BD * DE {DC = BD}
On solving (ii) & (iii), we get
⇒ AB² + AC² = AD² + BD² + 2BD * DE + AD² + BD² - 2BD * DE
= AD² + BD² + AD² + BD²
= 2(AD² + BD)²
Hence proved.!
