in triangle ABC, D is the midpoint of BC and AE is perpendicular to BC. if AC>AB. show that AB2 =AD2-BC2 * DE + 1/4
in triangle ABC D is the midpoint of BC and AE is perpendicular to BC . If AC>AB . Show that AB2=AD2-BC*DE+14BC2 .
Answers
Given:
i) ABC is a triangle, in which AC > AB; this is only for the inference that the altitude
from A on to BC lies towards B, that is point 'E' lies between B and D; this is the most
important inference for proving the given result.
ii) D is the midpoint of BC
iii) AE perpendicular BC
Required to prove: AB^2=AD^2-BC*DE+1/4BC^2
PROOF: [for easy convenience, let me denote all squares by * and multiplication by 'x']
1) Since AE is perpendicular to BC, ABE is a right triangle with angle E = 90 deg.
2) Hence applying Pythagoras theorem we have, AB* = AE* + BE*
3) In similar way from the right triangle, ADE, AD* = AE* + DE*, => AE* = AD* - DE*
4) Substituting for AE* from step 3 in step 2, we have,
AB* = AD* - DE* + BE*
5) From the figure (Kindly make according to the description given in data),
BE = BD - DE
6) Substituting this in (4), AB* = AD* - DE* + (BD-DE)*
7) Expanding, AB* = AD* - DE* + BD* + DE* - 2 x BD x DE
8) Cancelling -DE* and +DE* and substituting BD = (1/2)BC
{Since, D is mid point of BC}
we get, AB* = AD* + (1/4)BC* - BC x DE
Thus it is proved that "AB^2=AD^2-BC*DE+(1/4)BC^2"
Therefore, by Pythagoras theorem,
AB 2= AE2 + BE2 --------------- (1)
Now, AED is a right angled triangle.
Therefore,
=> AE2 + ED2= AD2 (Pythagoras theorem)
=> AE2 = AD2 - ED2 -------------- (2)
Now, BE = BD - ED ---------------- (3)
Substituting (2) and (3) in (1), we get,
AB2= AD2 - ED2 + (BD - ED)2
=AD2 – ED2 + BD2 -2BD*ED + ED2
= AD2 – ED2 + BD2 -2BD*ED + ED2
Now, BD = BC/2 (since, D is midpoint)
Therefore,
AB2= AD2 + BD2 -2BD*ED
=AD2 + (BC/2)2 – 2(BC/2)*ED
= AD2 + (BC/2)2 – 2(BC/2)*ED
= AD2 – BC*DE + ¼ BC2