Question

In isosceles triangle ∆ABC, BM is the median to the base AC . Point D is on BM . Prove the following triangle congruencies:∆ABD ≅ ∆CBD

Answer 1

Answer:

In isosceles triangle ABC, BM is the median to the base AC and Point D is on  BM as shown below in the figure;

Median of a triangle states that a line segment joining a vertex to the midpoint of the opposing side, bisecting it

M is the median of AC

then by definition;

AM = MC                 ......[1]

In  ΔAMD and  ΔDMC

AM = MC             [side]                [By [1]]

$\angle AMD = \angle DMC =90^{\circ}$    [Angle]

DM =DM       [Common side]      

Side-Angle-Side postulate(SAS) states that if  two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent.

Then, by SAS,

$\triangle AMD \cong \triangle DMC$

CPCT stands for Corresponding parts of congruent triangles are congruent

By CPCT,  

$AD = DC$       [Corresponding side]                   ......[2]

In  ΔABD and  ΔCBD

AB = BC   [Side]                      [By definition of isosceles triangle]

BD= BD   [common side]      

AD = DC [Side]                         [by [2]]

Side-Side-Side(SSS) postulate states that if three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent.

Therefore, by SSS theorem,

$\triangle ABD \cong \triangle CBD$    

Answer 2

Answer:

We know that triangle ABD and triangle CBD are congruent because of SAS.

Step-by-step explanation:

AB is congruent to BC because of the definition of an isosceles triangle

BD=BD because of the reflexive property

m<ABD=m<CBD because BM is the median of an isosceles triangle

Thus, triangle ABD is congruent to triangle CBD because of SAS