Question

$△\mathrm{ABC}$ is a right triangle with $\mathrm{AB}=\mathrm{AC}$. Bisector of $\angle A$ meets $\mathrm{BC}$ at $D$. Prove that $\mathrm{BC}=2\mathrm{AD}.$

Answer 1

Proving that $\mathrm{BC}=2\mathrm{AD}$:

Given:

A right angles triangle with $AB=AC$ bisector of $\angle A$ meets $BC$ at $D$.

To Prove:

$\mathrm{BC}=2\mathrm{AD}$

Proof

According to given details

From right$\mathrm{\Delta }\mathrm{ABC},$

$\mathrm{AB}=\mathrm{AC}$

Since, $\mathrm{BC}$ is hypotenuse

$\angle \mathrm{BAC}=90°$

From $\mathrm{\Delta CAD}$ and $\mathrm{\Delta BAD}$,

We have,

$\mathrm{AC}=\mathrm{AB}$ [given]

$\angle 1=\angle 2$ [AD is the bisector of ∠A ]

$\mathrm{AD}=\mathrm{AD}$ [Common side]

$\mathrm{\Delta }\mathrm{CAD}\cong \mathrm{\Delta }\mathrm{BAD}$ [ by $\mathrm{SAS}$ criteria ]

$\therefore \mathrm{CD}=\mathrm{BD}$ [ by$\mathrm{CPCT}$]

Since, Mid-point of hypotenuse of a right triangle is equidistant from the $3$ vertices of a triangle.

$AD=\mathrm{BD}=\mathrm{CD}\dots \left(1\right)$

Now, $\mathrm{BC}=\mathrm{BD}+\mathrm{CD}$ [Using eq.( $1$)]

$\mathrm{BC}=2\mathrm{AD}$

Hence, we proved that $\mathrm{BC}=2\mathrm{AD}$