Question

In a triangle ABC , the perpendicular bisector of AC meet AB at D .Prove that AB is equals to BD + DC

Answer 1

you can use SAS congruency in AED and CED then AD = CD
AB=AD+DB
AB=CD+DB proved

Answer 2

Given:

ABC is a triangle  

ED is the perpendicular bisector of AC

To find:

AB = BD + DC

Solution:

Since ED is the perpendicular bisector of AC so it will bisect AC equally at E

∴ AE = CE ...... (i)

Let's consider ΔAED and ΔCED, we have

∠DEA = ∠DEC = 90° ..... [∵ ED is perpendicular bisector of AC]

AE = CE ..... [from (i)]

ED = ED .... [common side]

∴ ΔAED ≅ ΔCED ..... [by SAS congruence]

According to the theorem of the Corresponding Parts of Congruent Triangles, we know, that if two triangles are congruent to each other then the corresponding angles and sides of the triangles are also congruent.

∴ AD = DC ..... (ii)

Now,

Referring to the given figure we get

AB = AD + BD

∵ from (ii), we have AD = DC

∴ AB = DC + BD

⇒ $\boxed{\bold{AB\: =\: BD\: +\: DC}}$

Hence Proved

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