Leila Jennings

Answered

2022-07-16

Question of Triangle of Geometry
In triangle ABC,we have $AB>AC$. If A' is the mid point of BC, AD is the altitude through A and if the internal and external bisectors of angle A meet at BC at X and X' respectively ,prove that ${A}^{\prime }D=\left({c}^{2}-{b}^{2}\right)/2a$
I have tried to use obtuse angle theorem to find height then solve the question but stuck on double variable.

Answer & Explanation

bulgarum87

Expert

2022-07-17Added 15 answers

Step 1
Since $c>b$, we obtain $\gamma >\beta ,$, which says that $\mathrm{\angle }ABC$ is an acute angle.
Thus, $BD=c\mathrm{cos}\beta =\frac{c\left({a}^{2}+{c}^{2}-{b}^{2}\right)}{2ac}=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2a}.$
Step 2
Also, me see that $BD>B{A}^{\prime }$ because $c>b$.
Thus, ${A}^{\prime }D=BD-B{A}^{\prime }=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2a}-\frac{a}{2}=\frac{{c}^{2}-{b}^{2}}{2a}.$

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