in triangle ABC ,angle b = 90 degree,D and E are points on side AC such that AD=DE=EC. prove that BD^2+ BE^2= 5DE^2.
Answers
Answer:
Step-by-step explanation:
Given information: ΔABC is a right angled triangle, ∠B=90° and D is the midpoint of BC.
To prove :
Proof :
According to the Pythagoras theorem, in a right angles triangle
Use Pythagoras theorem in triangle ABC....1
Use Pythagoras theorem in triangle ABD.
Equating (1) and (2), we get..... (2)
Add on both sides.
D is the midpoint of BC,
(BD=CD) (BC=2CD)
On further simplification, we get
Hence proved.
Answer:
Step-by-step explanation:
In Δ ABC, AB² + BC² = AC²
Given ∠B = 90°
Therefore, ∠A + ∠B + ∠C = 180°
∠B + ∠C = 90°
∠B = ∠C = 45°
Therefore, AB = AC
Now in Δ ABD and Δ DBE,
∠ abd = ∠ dbe = 30°
∠ADB = ∠ DEB = 105°
AD = DE
Therefore by AAS similarity criterion, ΔABD is similar to Δ DBE
Therefore, AB is similar to BD ........................ (1)
Similarly, Δ BEC is similar to ΔBDE
Therefore, BC is similar to BE ........................... (2)
Now use (1) and (2) in AB² + BC² = AC²
BD² + BE² = AC²
or, BD² + BE² = 3 DE² ( given that AD = DE = EC)
Hence proved.
