Question

If two isosceles triangles have a common base, then prove that the line joining their vertices bisects them at right angles.

Answer 1

Two issosceles triangles ABC and DBC having the common base BC such that AB = AC and DB = DC.
To prove : AD (or AD produced) bisects BC at right angle.


Proof : In $△$ABD and $△$ACD, we have
AB = AC [Given]
BD = CD [Given]
AD = AD [Common side]

So, by SSS criterion of congruence
$△$ABD $\cong △$ACD
$\Rightarrow \angle 1=\angle 2$.......(1)
[Corresponding parts of Congruent triangles are equal]
Now, in $△$'s ABE and ACE, we have
AB = AC [Given]
$\angle 1=\angle 2$ [From (1)]
and, AE = AE [Common side]
So, by SAS criterion of congruence,
$△ABE\cong △ACE\Rightarrow BE=CEand\angle 3=\angle 4$
[Corresponding parts of Congruent triangles are equal]
But, $\angle 3+\angle 4={180}^{\circ }$
[Sum of the angles of a linear pair is ${180}^{\circ }$]
$\Rightarrow 2\angle 3={180}^{\circ }[\angle 3=\angle 4]\Rightarrow \angle 3={90}^{\circ }\therefore \angle 3=\angle 4={90}^{\circ }$
Hence, AD bisects BC at right angles.