Properties and Solution of Triangles
11.0 Length of Angle Bisectors, Medians and Altitudes
11.0 Length of Angle Bisectors, Medians and Altitudes
$(i)$ The length of angle bisector from the angle $A$ is given as $${\beta _a} = \frac{{2bc\cos \frac{A}{2}}}{{b + c}}$$
Proof: From the figure,we have $$\begin{equation} \begin{aligned} BD = \frac{{ac}}{{b + c}} \\ DC = \frac{{ab}}{{b + c}} \\\end{aligned} \end{equation} $$
From the definition of angle bisector, we have $$\angle BAD \cong \angle FAC$$
And we know that angle in the same segment of circle are equal so we have $$\angle ABD \cong \angle AFC$$
Now triangles with two angles equal are similar so we have $$\Delta ABD \sim \Delta AFC$$ so$$\begin{equation} \begin{aligned} \frac{c}{d} = \frac{{AF}}{b} \\ \frac{c}{d} = \frac{{d + DF}}{b} \\\end{aligned} \end{equation} $$
Using intersecting chord theorem, we have $$BD \times DC = d \times DF$$
therefore putting the values we get $$\begin{equation} \begin{aligned} \frac{c}{d} = \frac{{d + \frac{{BD \times DC}}{d}}}{b} \\ bc = {d^2} + (BD \times DC) \\ {d^2} = bc - (BD \times DC) \\ {d^2} = bc - \left( {\frac{{ac}}{{b + c}} \times \frac{{ab}}{{b + c}}} \right) \\ {d^2} = \frac{{bc}}{{{{(b + c)}^2}}}\left( {{{(b + c)}^2} - {a^2}} \right) \\\end{aligned} \end{equation} $$
Now$${d^2} = \frac{{bc}}{{{{(b + c)}^2}}}\left( {{b^2} + {c^2} + 2bc - {a^2}} \right)$$
Now using cosine rule $$\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$$ in the above expression , we get $$\begin{equation} \begin{aligned} {d^2} = \frac{{bc}}{{{{(b + c)}^2}}}\left( {2bc\cos A + 2bc} \right) \\ {d^2} = \frac{{2{{(bc)}^2}}}{{{{(b + c)}^2}}}\left( {1 + \cos A} \right) \\ {d^2} = \frac{{2{{(bc)}^2}}}{{{{(b + c)}^2}}}(2{\cos ^2}\frac{A}{2}) \\\end{aligned} \end{equation} $$
Now take square root we get $$d = \frac{{2bc\cos \frac{A}{2}}}{{(b + c)}}$$
Replacing $d$ by ${\beta _a}$ , we have $${\beta _a} = \frac{{2bc\cos \frac{A}{2}}}{{b + c}}$$
$(ii)$ Length of median from the vertex $A$ is given as $${m_a} = \frac{1}{2}\sqrt {2{b^2} + 2{c^2} - {a^2}} $$
Proof: Figure shows the triangle $ABC$ with the sides $a=BC$, $b=AC$ and $c=AB$ and $m=CD$ is the median drawn to the side $c$. Let $d$ be the length of the segment $AD$ and $e$ be the length of the segment $DB$ .
The median divides the side $AB$. Let $\delta $ and $\phi $ are the angles of the intersection of the median $CD$ to the side $AB$.
In triangle $ADC$, apply law of cosines to find the length of $AC$, we have $${b^2} = {d^2} + {m^2} - 2dm\cos \phi ...(i)$$
Similarly, in triangle $BDC$, apply law of cosines to find the length of $BC$, we have $${a^2} = {e^2} + {m^2} - 2em\cos \delta ...(ii)$$
since $CD$ is the median so $d=e$, and also $\delta $ and $\phi $ angles are supplementary so we have $$\cos \delta = - \cos \phi $$
Adding $(i)$ and $(ii)$, we get
$${b^2} + {a^2} = {d^2} + {e^2} + {m^2} + {m^2} - 2dm\cos \phi - 2em\cos \phi $$
or, we have $${b^2} + {a^2} = {d^2} + {e^2} + 2{m^2}$$
Substitute $$\begin{equation} \begin{aligned} d = \frac{c}{2} \\ e = \frac{c}{2} \\\end{aligned} \end{equation} $$ as $CD$ is the median.
We get $$\begin{equation} \begin{aligned} {m^2} = \frac{{\left( {2{b^2} + 2{a^2} - {c^2}} \right)}}{4} \\ m = \frac{1}{2}\sqrt {2{b^2} + 2{a^2} - {c^2}} \\\end{aligned} \end{equation} $$
or replace $m$ by ${m_a}$, we have $${m_a} = \frac{1}{2}\sqrt {2{b^2} + 2{a^2} - {c^2}} $$
$(iii)$ Length of altitude from the angle $A$ is given as $${A_a} = \frac{{2\Delta }}{a}$$
Proof: Consider a triangle $ABC$ in which altitude is drawn from vertex $A$ on side $BC$ which meets at a point $D$. So, we need to find the length of $AD$.
from figure, we can write as $$sinB = \frac{{AD}}{c}$$
or,$$AD = c\sin B$$
Multiplying and dividing by $2a$, we get
or we can write it as $$\begin{equation} \begin{aligned} = \frac{{2a}}{{2a}}c\sin B \\ = \frac{2}{a} \times \left( {\frac{1}{2}ac\sin B} \right) \\ = \frac{2}{a}\Delta \\ = \frac{{2\Delta }}{a} \\\end{aligned} \end{equation} $$
Question 23. In a triangle with sides $a$, $b$ and$c$, the lengths of the medians are ${m_a}$, ${m_b}$, ${m_c}$ drawn to the sides $a$, $b$ and$c$ respectively, then prove that $${m_a}^2 + {m_b}^2 + {m_c}^2 = \frac{3}{4}\left( {{a^2} + {b^2} + {c^2}} \right)$$
Solution: As we know that length of the median drawn from vertex $A$ is given as $${m_a} = \frac{1}{2}\sqrt {2{b^2} + 2{c^2} - {a^2}} $$
So squaring both sides, we get $${m_a}^2 = \frac{{\left( {2{b^2} + 2{c^2} - {a^2}} \right)}}{4}...(i)$$
Similarly we can write down the length of median drawn from vertex $B$ as
$${m_b}^2 = \frac{{\left( {2{c^2} + 2{a^2} - {b^2}} \right)}}{4}...(ii)$$
and the length of median drawn from vertex $C$ as $${m_c}^2 = \frac{{\left( {2{a^2} + 2{b^2} - {c^2}} \right)}}{4}...(iii)$$
Adding $(i)$, $(ii)$ and $(iii)$, we get
$${m_a}^2 + {m_b}^2 + {m_c}^2 = \frac{{\left( {2{b^2} + 2{c^2} - {a^2}} \right)}}{4} + \frac{{\left( {2{c^2} + 2{a^2} - {b^2}} \right)}}{4} + \frac{{\left( {2{a^2} + 2{b^2} - {c^2}} \right)}}{4}$$
$$\begin{equation} \begin{aligned} \frac{1}{4}\left( {3{a^2} + 3{b^2} + 3{c^2}} \right) \\ \frac{3}{4}\left( {{a^2} + {b^2} + {c^2}} \right) \\\end{aligned} \end{equation} $$
Note: $${m_a}^2 + {m_b}^2 + {m_c}^2 = \frac{3}{4}\left( {{a^2} + {b^2} + {c^2}} \right)$$
Question 24. In the isosceles triangle the lateral side has the length of $4$. The median drawn to the lateral side has the length of $3$. Find the length of the base of the triangle.
Solution: Let $x$ be the length of the base of a triangle.
Now apply the formula of the length of median $${m_b} = \frac{{\sqrt {\left( {2{c^2} + 2{a^2} - {b^2}} \right)} }}{2}$$, we get
$$3 = \frac{1}{2}\sqrt {2 \times {4^2} + 2 \times {x^2} - {4^2}} $$
Squaring both sides,
$$\begin{equation} \begin{aligned} 9 = \frac{1}{4}\left( {32 + 2{x^2} - 16} \right) \\ 36 = 16 + 2{x^2} \\ 20 = 2{x^2} \\ {x^2} = 10 \\ x = \sqrt {10} \\\end{aligned} \end{equation} $$
So the length of base of triangle is $\sqrt {10} $.
Question 25. $AD$ is the median of triangle $ABC$. If $AE$ and $AF$ are medians of the traingle $ABD$ and $ADC$ respectively and $AD = {m_1}$, $AE = {m_2}$ and $AF = {m_3}$, the prove that $${m_2}^2 + {m_3}^2 - 2{m_1}^2 = \frac{{{a^2}}}{8}$$
Solution: In $\Delta ABC$
$$A{D^2} = \frac{1}{4}\left( {2{b^2} + 2{c^2} - {a^2}} \right) = {m_1}^2...(i)$$
Similarly in $\Delta ABD$
$$A{E^2} = \frac{1}{4}\left( {2{c^2} + 2A{D^2} - \frac{{{a^2}}}{4}} \right) = {m_2}^2...(ii)$$
Similarly in $\Delta ADC$
$$A{F^2} = \frac{1}{4}\left( {2{b^2} + 2A{D^2} - \frac{{{a^2}}}{4}} \right) = {m_3}^2...(iii)$$
Adding equations $(ii)$ and $(iii)$, we get $$\begin{equation} \begin{aligned} {m_2}^2 + {m_3}^2 = \frac{1}{4}\left( {2{c^2} + 2A{D^2} - \frac{{{a^2}}}{4}} \right) + \frac{1}{4}\left( {2{b^2} + 2A{D^2} - \frac{{{a^2}}}{4}} \right) \\ = \frac{1}{4}\left( {2{c^2} + 4A{D^2} + 2{b^2} - \frac{{{a^2}}}{2}} \right) \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} = A{D^2} + \frac{1}{4}\left( {2{c^2} + 2{b^2} - \frac{{{a^2}}}{2}} \right) \\ = A{D^2} + \frac{1}{4}\left( {2{c^2} + 2{b^2} - {a^2} + \frac{{{a^2}}}{2}} \right) \\ = A{D^2} + \frac{1}{4}\left( {2{c^2} + 2{b^2} - {a^2}} \right) + \frac{{{a^2}}}{8} \\ = A{D^2} + A{D^2} + \frac{{{a^2}}}{8} \\ = 2A{D^2} + \frac{{{a^2}}}{8} \\\end{aligned} \end{equation} $$
As we know that $$A{D^2} = {m_1}^2$$
therefore, $$\begin{equation} \begin{aligned} = 2{m_1}^2 + \frac{{{a^2}}}{8} \\ {m_2}^2 + {m_3}^2 = 2{m_1}^2 + \frac{{{a^2}}}{8} \\ {m_2}^2 + {m_3}^2 - 2{m_1}^2 = \frac{{{a^2}}}{8} \\\end{aligned} \end{equation} $$
Question 26. In a $\Delta ABC$ if $a=5cm$, $b=4cm$ and $c=3cm$. $'G'$ is the centroid of the triangle, then find the circumradius of $\Delta ABC$.
Solution: We know that the centroid divides the median in the ration of $2:1$ so the length $AG$ can be calculated as
$$AG = \frac{2}{3} \times \frac{1}{2}\sqrt {\left( {2{b^2} + 2{c^2} - {a^2}} \right)} $$As length of median drawn from vertex A is given as $$\frac{1}{2}\sqrt {\left( {2{b^2} + 2{c^2} - {a^2}} \right)} $$
$$\begin{equation} \begin{aligned} = \frac{2}{3} \times \frac{1}{2}\sqrt {\left( {2 \times {4^2} + 2 \times {3^2} - {5^2}} \right)} \\ = \frac{1}{3}\sqrt {32 + 18 - 25} \\ = \frac{5}{3} \\\end{aligned} \end{equation} $$
Therefore, we have $$AG = \frac{5}{3}$$
Similarly we can find the length $BG$ so $$\begin{equation} \begin{aligned} BG = \frac{2}{3} \times \frac{1}{2}\sqrt {\left( {2{a^2} + 2{c^2} - {b^2}} \right)} \\ = \frac{2}{3} \times \frac{1}{2}\sqrt {\left( {2 \times {5^2} + 2 \times {3^2} - {4^2}} \right)} \\ = \frac{1}{3}\sqrt {50 + 18 - 16} \\ = \frac{{2\sqrt {13} }}{3} \\\end{aligned} \end{equation} $$
$$BG = \frac{{2\sqrt {13} }}{3}$$
Now in $\Delta GAB$, we have $$AG = \frac{5}{3}$$
$$BG = \frac{{2\sqrt {13} }}{3}$$
$$AB=3$$
Circumradius is given as $$R = \frac{{abc}}{{4\Delta }}$$ where $$\begin{equation} \begin{aligned} \Delta = \sqrt {s(s - a)(s - b)(s - c)} \\ s = \frac{{a + b + c}}{2} \\\end{aligned} \end{equation} $$
so first find $$\begin{equation} \begin{aligned} s = \frac{{a + b + c}}{2} \\ = \frac{{\frac{5}{3} + \frac{{2\sqrt {13} }}{3} + 3}}{2} \\ = \frac{{14 + 2\sqrt {13} }}{6} \\ = \frac{{7 + \sqrt {13} }}{3} \\\end{aligned} \end{equation} $$
Now find area
$$\begin{equation} \begin{aligned} \Delta = \sqrt {\left( {\frac{{7 + \sqrt {13} }}{3}} \right) \times \left( {\frac{{7 + \sqrt {13} }}{3} - \frac{5}{3}} \right) \times \left( {\frac{{7 + \sqrt {13} }}{3} - \frac{{2\sqrt {13} }}{3}} \right) \times \left( {\frac{{7 + \sqrt {13} }}{3} - 3} \right)} \\ \Delta = \sqrt {\left( {\frac{{7 + \sqrt {13} }}{3}} \right) \times \left( {\frac{{7 + \sqrt {13} }}{3} - \frac{5}{3}} \right) \times \left( {\frac{{7 + \sqrt {13} }}{3} - \frac{{2\sqrt {13} }}{3}} \right) \times \left( {\frac{{7 + \sqrt {13} }}{3} - 3} \right)} \\ \Delta = \sqrt {\left( {\frac{{7 + \sqrt {13} }}{3}} \right)\left( {\frac{{7 - \sqrt {13} }}{3}} \right)\left( {\frac{{2 + \sqrt {13} }}{3}} \right)\left( {\frac{{ - 2 + \sqrt {13} }}{3}} \right)} \\ \Delta = \sqrt {\left( {\frac{{49 - 13}}{9}} \right)\left( {\frac{{13 - 4}}{9}} \right)} = \sqrt {\frac{{36}}{9}} = 2 \\\end{aligned} \end{equation} $$
Substituting the values in the formula of circumradius, we get $$\begin{equation} \begin{aligned} R = \frac{{abc}}{{4\Delta }} \\ = \frac{{\frac{5}{3} \times \frac{{2\sqrt {13} }}{3} \times 3}}{{4 \times 2}} \\ = \frac{5}{{12}}\sqrt {13} \\\end{aligned} \end{equation} $$
Therefore, circumradius of triangle $GAB$ is $$\frac{5}{{12}}\sqrt {13} $$
