In the given figure, $AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that $\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \times \mathrm{DM}+(\frac{\mathrm{BC}}{2})^2$ "
Given:
$AD$ is a median of a triangle $ABC$ and $AM \perp BC$.
To do:
We have to prove that $\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \times \mathrm{DM}+(\frac{\mathrm{BC}}{2})^2$
Solution:
In $\triangle \mathrm{AMB}$, by Pythagoras theorem,
$\mathrm{AB}^{2}=\mathrm{AM}^{2}+\mathrm{BM}^{2}$
$\mathrm{AB}^{2}=\mathrm{AM}^{2}+(\mathrm{BD}-\mathrm{MD})^{2}$
$\mathrm{AB}^{2}=\mathrm{AM}^{2}+(\frac{\mathrm{BC}}{2}-\mathrm{MD})^{2}$
$\mathrm{AB}^{2}=\mathrm{AM}^{2}+\mathrm{MD}^{2}+(\frac{\mathrm{BC}}{2})^{2}-\frac{2 \mathrm{BC}}{2} \times \mathrm{MD}$
$\mathrm{AB}^{2}=(\mathrm{AM}^{2}+\mathrm{MD}^{2})+(\frac{\mathrm{BC}}{2})^{2}-\mathrm{BC} \times \mathrm{MD}$
$\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \times \mathrm{MD}+(\frac{\mathrm{BC}}{2})^{2}$
Hence proved.
Related Articles In the given figure, $AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that(i)$\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{BC} \times \mathrm{DM}+(\frac{\mathrm{BC}}{2})^{2}$(ii) $\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \times \mathrm{DM}+(\frac{\mathrm{BC}}{2})^2$(iii) $\mathrm{AC}^{2}+\mathrm{AB}^{2}=2 \mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}^{2}$"
In the given figure, $AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that$\mathrm{AC}^{2}+\mathrm{AB}^{2}=2 \mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}^{2}$"
In \( \triangle \mathrm{ABC}, \mathrm{AD} \) is a median. If \( \mathrm{AB}=8, \mathrm{AC}=15 \) and \( \mathrm{AD}=8.5 \), find \( \mathrm{BC} \).
Choose the correct answer from the given four options:In the figure below, \( \angle \mathrm{BAC}=90^{\circ} \) and \( \mathrm{AD} \perp \mathrm{BC} \). Then,(A) \( \mathrm{BD} \cdot \mathrm{CD}=\mathrm{BC}^{2} \)(B) \( \mathrm{AB}
In a quadrilateral \( \mathrm{ABCD}, \angle \mathrm{A}+\angle \mathrm{D}=90^{\circ} \). Prove that \( \mathrm{AC}^{2}+\mathrm{BD}^{2}=\mathrm{AD}^{2}+\mathrm{BC}^{2} \) [Hint: Produce \( \mathrm{AB} \) and DC to meet at E]
In \( \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} \) and \( \mathrm{BM} \) is an altitude. If \( \mathrm{AM}=2 x^{2} \) and \( \mathrm{CM}=8 x^{2} \), find \( \mathrm{BM} \), AB and \( \mathrm{BC} \).
\( \mathrm{AD} \) is an altitude of an isosceles triangle \( \mathrm{ABC} \) in which \( \mathrm{AB}=\mathrm{AC} \). Show that(i) ADbisects BC(ii) \( \mathrm{AD} \) bisects \( \angle \mathrm{A} \).
In an equilateral $\triangle ABC$, $AD \perp BC$, prove that $AD^2=3BD^2$.
In \( \triangle \mathrm{ABC}, \mathrm{AD} \) is the perpendicular bisector of \( \mathrm{BC} \) (see Fig. 7.30). Show that \( \triangle \mathrm{ABC} \) is an isosceles triangle in which \( \mathrm{AB}=\mathrm{AC} \)."\n
In \( \triangle \mathrm{ABC}, \mathrm{AD} \) is a median. If \( \mathrm{AB}=18 \mathrm{~cm} \). \( \mathrm{AC}=14 \mathrm{~cm} \) and \( \mathrm{AD}=14 \mathrm{~cm} \), find the perimeter of \( \triangle \mathrm{ABC} \).
In \( \triangle \mathrm{ABC}, \mathrm{M} \) and \( \mathrm{N} \) are points of \( \mathrm{AB} \) and \( \mathrm{AC} \) respectively and \( \mathrm{MN} \| \mathrm{BC} \). If \( \mathrm{AM}=x \), \( \mathrm{MB}=x-2, \mathrm{AN}=x+2 \) and \( \mathrm{NC}=x-1 \), find the value of \( x \).
In \( \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} \). If \( \mathrm{AC}-\mathrm{BC}=4 \) and \( \mathrm{BC}-\mathrm{AB}=4 \), find all the three sides of \( \triangle \mathrm{ABC} \).
In a \( \Delta \mathrm{PQR}, \mathrm{PR}^{2}-\mathrm{PQ}^{2}=\mathrm{QR}^{2} \) and \( \mathrm{M} \) is a point on side \( \mathrm{PR} \) such that \( \mathrm{QM} \perp \mathrm{PR} \).Prove that \( \mathrm{QM}^{2}=\mathrm{PM} \times \mathrm{MR} \).
In \( \triangle \mathrm{ABC} \) and \( \triangle \mathrm{DEF}, \mathrm{AB}=\mathrm{DE}, \mathrm{AB} \| \mathrm{DE}, \mathrm{BC}=\mathrm{EF} \) and \( \mathrm{BC} \| EF \). Vertices \( \mathrm{A}, \mathrm{B} \) and \( \mathrm{C} \) are joined to vertices D, E and F respectively (see below figure).Show that(i) quadrilateral ABED is a parallelogram(ii) quadrilateral \( \mathrm{BEFC} \) is a parallelogram(iii) \( \mathrm{AD} \| \mathrm{CF} \) and \( \mathrm{AD}=\mathrm{CF} \)(iv) quadrilateral ACFD is a parallelogram(v) \( \mathrm{AC}=\mathrm{DF} \)(vi) \( \triangle \mathrm{ABC} \equiv \triangle \mathrm{DEF} \)."
\( \mathrm{ABC} \) is an isosceles triangle with \( \mathrm{AB}=\mathrm{AC} \). Draw \( \mathrm{AP} \perp \mathrm{BC} \) to show that \( \angle \mathrm{B}=\angle \mathrm{C} \).
Kickstart Your Career
Get certified by completing the course
Get Started