Question

in a triangle abc ac >ab. d is the midpoint of bc and AD perpendicular bisector of BC. prove that (1) AC²=AD²+BC. DE +¼BC² (2) AB²=AD² - BC. DC+¼BC²​

Answer 1

A triangle ABC in which AB = AC and D is any point

in BC.

To Prove:

AB2-AD2 = BD.CD

Const: Draw AEI BC

Proof: In AABE and AACE, we have AB = AC [given]

AE = AE [common] and ZEB = ZAC [90] Therefore, by using RH congruent condition

AABE - AACE

BE = CE

In right triangle ABE.

AB2 = AE2 + BE2 .(i) [Using Pythagoras theorem)

In right triangle ADE, AD2 = AE? + DE?

[Using Pythagoras theorem]

Subtracting (ii) from (i), we get

AB2 - AD2 = (AE2 + BE?) - (AE2 + DE2) AB2-AD2 = AE2 + BE? - AE? - DE?

AB2 - AD2 = BE2 - DE2 But BE = CE [Proved above]

= AB? - AD2 (BE + DE) (BE - DE)

> AB2 - AD2 = (CE + DE) (BE - DE) = CD.BD

> AB - AD = BD.CD Hence Proved.

HOPE THIS HELPS YOU SOMEHOW......!!!

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