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 ${A}^{\prime}D=({c}^{2}-{b}^{2})/2a$

I have tried to use obtuse angle theorem to find height then solve the question but stuck on double variable.

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 ${A}^{\prime}D=({c}^{2}-{b}^{2})/2a$

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, $BD=c\mathrm{cos}\beta =\frac{c({a}^{2}+{c}^{2}-{b}^{2})}{2ac}=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2a}.$

Step 2

Also, me see that $BD>B{A}^{\prime}$ because $c>b$.

Thus, ${A}^{\prime}D=BD-B{A}^{\prime}=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2a}-\frac{a}{2}=\frac{{c}^{2}-{b}^{2}}{2a}.$

Since $c>b$, we obtain $\gamma >\beta ,$, which says that $\mathrm{\angle}ABC$ is an acute angle.

Thus, $BD=c\mathrm{cos}\beta =\frac{c({a}^{2}+{c}^{2}-{b}^{2})}{2ac}=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2a}.$

Step 2

Also, me see that $BD>B{A}^{\prime}$ because $c>b$.

Thus, ${A}^{\prime}D=BD-B{A}^{\prime}=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2a}-\frac{a}{2}=\frac{{c}^{2}-{b}^{2}}{2a}.$

Most Popular Questions