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(N/A) Given: $	riangle ABC$ with median $AD$ to side $BC$.To prove: $AB + AC > 2AD$.Construction: Produce $AD$ to $E$ such that $DE = AD$ and join $E

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(N/A) Given: $	riangle ABC$ with median $AD$ to side $BC$.To prove: $AB + AC > 2AD$.Construction: Produce $AD$ to $E$ such that $DE = AD$ and join $EC$.Proof: In $	riangle ADB$ and $	riangle EDC$:$AD = ED$ (By construction)$\angle 1 = \angle 2$ (Vertically opposite angles)$DB = DC$ ($AD$ is the median)By $SAS$ congruence criterion,$	riangle ADB \cong 	riangle EDC$.Therefore,$AB = EC$ $(CPCT)$.Now,in $	riangle AEC$,the sum of any two sides is greater than the third side:$AC + CE > AE$Since $AE = AD + DE = AD + AD = 2AD$ and $CE = AB$:$AC + AB > 2AD$.Hence,it is proved that the sum of two sides is greater than twice the median to the third side.

Prove that the sum of any two sides of a triangle is greater than twice the median with respect to the third side.

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State whether each of the following statements is true or false:
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Prove that the sum of any two sides of a triangle is greater than twice the median with respect to the third side.

Similar Questions

State whether each of the following statements is true or false:$(1)$ In $\Delta XYZ$,if $XY > XZ$,then $\angle Z > \angle Y$.$(2)$ In $\Delta ABC$ and $\Delta PQR$,if $\frac{AB}{PR} = \frac{BC}{QP} = \frac{CA}{RQ} = 1$,then $\Delta ABC \cong \Delta RPQ$.

In the given figure,$AB = AC$ and $BP = CQ$. Prove that $\Delta APQ$ is an isosceles triangle.

Which of the following is not a criterion for congruence of triangles?

In a triangle $ABC$,$D$ is the mid-point of side $AC$ such that $BD = \frac{1}{2} AC$. Show that $\angle ABC$ is a right angle.

Prove that the sum of any two sides of a triangle is greater than twice the median with respect to the third side.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

State whether each of the following statements is true or false:
$(1)$ In $\Delta XYZ$,if $XY > XZ$,then $\angle Z > \angle Y$.
$(2)$ In $\Delta ABC$ and $\Delta PQR$,if $\frac{AB}{PR} = \frac{BC}{QP} = \frac{CA}{RQ} = 1$,then $\Delta ABC \cong \Delta RPQ$.