In $\Delta ABC$, prove $A'D=\frac{(c^2- b^2)}{2a}$ with $A'$ be the midpoint of $BC$, $AB>AC$, and other Information given.

Question

Problem

In $\Delta ABC$, we have $AB>AC$.

• Let $A'$ be the midpoint of $BC$
• $D$ be the foot of altitude from $A$ to $BC$
• Internal and external Angle Bisectors of $\angle A$ meet $BC$ at $X$ and $X'$ respectively.

Prove that

1. $A'X = \frac{a(c-b)}{2(c+b)}$
2. $A'X' = \frac{a(c+b)}{2(c-b)}$
3. $A'D=\frac{(c^2- b^2)}{2a}$ where $a, b, c$ are the sides $BC, CA, AB$ as used commonly.

[Question Taken From Challenge and Thrill of Pre-College Mathematics]

My Approach
I could solve part 1. and 2. using the angle bisector theorem a few times, and with some pre-requisite knowledge, which were in-fact to be proven in questions in the exercise preceding this one:

• $XC= \frac{ab}{(b+c)}$, consequently $XB= \frac{ac}{(b+c)}$
• $CX'= \frac{ac}{(c-b)}$, consequently $BX'= \frac{ab}{(c-b)}$
• $XX'= \frac{2abc}{(c^2- b^2)}$

While I could prove the given formulas separately, I have no idea how to solve part 3.. A hint to help me start off would be appreciated.

Answer 1

Let $CD=x$. Apply Pythagorean Theorem on $\triangle ADC$ and $\triangle ADB$ to get $b^{2}-x^{2}=c^{2}-(a-x)^{2}$. Solve for $x$ here( the $x^{2}$-s will cancel out).

Now, $A'D=A'C-CD=\frac{a}{2}-x$

Answer 2

If you are allowed to use law of cosines, then it is very straightforward.

$A'D = c \cos B - \frac{a}{2}$

and $\displaystyle \cos B = \frac{a^2 + c^2 - b^2}{2ac}$.

That is all it takes.